first commit
commit
5c4be8335a
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@ -0,0 +1,12 @@
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'''
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【程序1】
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题目:有1、2、3、4个数字,能组成多少个互不相同且无重复数字的三位数?都是多少?
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1.程序分析:可填在百位、十位、个位的数字都是1、2、3、4。组成所有的排列后再去
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掉不满足条件的排列。
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2.程序源代码:
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'''
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for i in range(1,5):
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for j in range(1,5):
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for k in range(1,5):
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if( i != k ) and (i != j) and (j != k):
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print i,j,k
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'''
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【程序2】
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题目:企业发放的奖金根据利润提成。利润(I)低于或等于10万元时,奖金可提10%;利润高
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于10万元,低于20万元时,低于10万元的部分按10%提成,高于10万元的部分,可可提
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成7.5%;20万到40万之间时,高于20万元的部分,可提成5%;40万到60万之间时高于
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40万元的部分,可提成3%;60万到100万之间时,高于60万元的部分,可提成1.5%,高于
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100万元时,超过100万元的部分按1%提成,从键盘输入当月利润I,求应发放奖金总数?
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1.程序分析:请利用数轴来分界,定位。注意定义时需把奖金定义成长整型。
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2.程序源代码:
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'''
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bonus1 = 100000 * 0.1
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bonus2 = bonus1 + 100000 * 0.500075
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bonus4 = bonus2 + 200000 * 0.5
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bonus6 = bonus4 + 200000 * 0.3
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bonus10 = bonus6 + 400000 * 0.15
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i = int(raw_input('input gain:\n'))
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if i <= 100000:
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bonus = i * 0.1
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elif i <= 200000:
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bonus = bonus1 + (i - 100000) * 0.075
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elif i <= 400000:
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bonus = bonus2 + (i - 200000) * 0.05
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elif i <= 600000:
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bonus = bonus4 + (i - 400000) * 0.03
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elif i <= 1000000:
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bonus = bonus6 + (i - 600000) * 0.015
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else:
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bonus = bonus10 + (i - 1000000) * 0.01
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print 'bonus = ',bonus
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'''
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【程序3】
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题目:一个整数,它加上100后是一个完全平方数,再加上168又是一个完全平方数,请问该数是多少?
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1.程序分析:在10万以内判断,先将该数加上100后再开方,再将该数加上268后再开方,如果开方后
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的结果满足如下条件,即是结果。请看具体分析:
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2.程序源代码:
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#include "math.h"
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main()
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{
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long int i,x,y,z;
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for (i=1;i<100000;i++)
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{ x=sqrt(i+100); /*x为加上100后开方后的结果*/
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y=sqrt(i+268); /*y为再加上168后开方后的结果*/
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if(x*x==i+100&&y*y==i+268)/*如果一个数的平方根的平方等于该数,这说明此数是完全平方数*/
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printf("\n%ld\n",i);
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}
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}
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'''
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import math
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for i in range(10000):
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#转化为整型值
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x = int(math.sqrt(i + 100))
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y = int(math.sqrt(i + 268))
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if(x * x == i + 100) and (y * y == i + 268):
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print i
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'''题目:输入某年某月某日,判断这一天是这一年的第几天?
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1.程序分析:以3月5日为例,应该先把前两个月的加起来,然后再加上5天即本年的第几天,特殊
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情况,闰年且输入月份大于3时需考虑多加一天。
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2.程序源代码:
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'''
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year = int(raw_input('year:\n'))
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month = int(raw_input('month:\n'))
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day = int(raw_input('day:\n'))
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months = (0,31,59,90,120,151,181,212,243,273,304,334)
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if 0 <= month <= 12:
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sum = months[month - 1]
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else:
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print 'data error'
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sum += day
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leap = 0
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if (year % 400 == 0) or ((year % 4 == 0) and (year % 100 != 0)):
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leap = 1
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if (leap == 1) and (month > 2):
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sum += 1
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print 'it is the %dth day.' % sum
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'''
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【程序5】
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题目:输入三个整数x,y,z,请把这三个数由小到大输出。
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1.程序分析:我们想办法把最小的数放到x上,先将x与y进行比较,如果x>y则将x与y的值进行交换,
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然后再用x与z进行比较,如果x>z则将x与z的值进行交换,这样能使x最小。
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2.程序源代码:
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'''
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l = []
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for i in range(3):
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x = int(raw_input('integer:\n'))
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l.append(x)
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l.sort()
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print l
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'''
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【程序6】
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题目:用*号输出字母C的图案。
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1.程序分析:可先用'*'号在纸上写出字母C,再分行输出。
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2.程序源代码:
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'''
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print 'Hello Python world!\n'
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print '*' * 10
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for i in range(5):
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print '* *'
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print '*' * 10
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print '*\n' * 6
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'''
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【程序7】
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题目:输出特殊图案,请在c环境中运行,看一看,Very Beautiful!
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1.程序分析:字符共有256个。不同字符,图形不一样。
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2.程序源代码:
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'''
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a = 176
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b = 219
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print chr(b),chr(a),chr(a),chr(a),chr(b)
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print chr(a),chr(b),chr(a),chr(b),chr(a)
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print chr(a),chr(a),chr(b),chr(a),chr(a)
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print chr(a),chr(b),chr(a),chr(b),chr(a)
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print chr(b),chr(a),chr(a),chr(a),chr(b)
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'''
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【程序8】
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题目:输出9*9口诀。
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1.程序分析:分行与列考虑,共9行9列,i控制行,j控制列。
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2.程序源代码:
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#include "stdio.h"
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main()
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{
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int i,j,result;
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printf("\n");
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for (i=1;i<10;i++)
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{ for(j=1;j<10;j++)
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{
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result=i*j;
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printf("%d*%d=%-3d",i,j,result);/*-3d表示左对齐,占3位*/
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}
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printf("\n");/*每一行后换行*/
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}
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}
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'''
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for i in range(1,10):
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for j in range(1,10):
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result = i * j
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print '%d * %d = % -3d' % (i,j,result)
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print ''
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'''
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【程序9】
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题目:要求输出国际象棋棋盘。
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1.程序分析:用i控制行,j来控制列,根据i+j的和的变化来控制输出黑方格,还是白方格。
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2.程序源代码:
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#include "stdio.h"
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main()
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{
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int i,j;
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for(i=0;i<8;i++)
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{
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for(j=0;j<8;j++)
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if((i+j)%2==0)
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printf("%c%c",219,219);
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else
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printf(" ");
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printf("\n");
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}
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}
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'''
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import sys
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for i in range(8):
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for j in range(8):
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if(i + j) % 2 == 0:
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sys.stdout.write(chr(219))
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sys.stdout.write(chr(219))
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else:
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sys.stdout.write(' ')
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print ''
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'''
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【程序10】
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题目:打印楼梯,同时在楼梯上方打印两个笑脸。
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1.程序分析:用i控制行,j来控制列,j根据i的变化来控制输出黑方格的个数。
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2.程序源代码:
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'''
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import sys
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sys.stdout.write(chr(1))
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sys.stdout.write(chr(1))
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print ''
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for i in range(1,11):
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for j in range(1,i):
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sys.stdout.write(chr(219))
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sys.stdout.write(chr(219))
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print ''
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'''
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【程序11】
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题目:古典问题:有一对兔子,从出生后第3个月起每个月都生一对兔子,小兔子长到第三个月
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后每个月又生一对兔子,假如兔子都不死,问每个月的兔子总数为多少?
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1.程序分析: 兔子的规律为数列1,1,2,3,5,8,13,21....
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2.程序源代码:
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main()
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{
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long f1,f2;
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int i;
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f1=f2=1;
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for(i=1;i<=20;i++)
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{ printf("%12ld %12ld",f1,f2);
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if(i%2==0) printf("\n");/*控制输出,每行四个*/
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f1=f1+f2; /*前两个月加起来赋值给第三个月*/
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f2=f1+f2; /*前两个月加起来赋值给第三个月*/
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}
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}
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'''
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f1 = 1
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f2 = 1
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for i in range(1,21):
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print '%12d %12d' % (f1,f2)
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if (i % 2) == 0:
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print ''
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f1 = f1 + f2
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f2 = f1 + f2
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'''
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【程序12】
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题目:判断101-200之间有多少个素数,并输出所有素数。
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1.程序分析:判断素数的方法:用一个数分别去除2到sqrt(这个数),如果能被整除,
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则表明此数不是素数,反之是素数。
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2.程序源代码:
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'''
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h = 0
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leap = 1
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from math import sqrt
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from sys import stdout
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for m in range(101,201):
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k = int(sqrt(m + 1))
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for i in range(2,k + 1):
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if m % i == 0:
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leap = 0
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break
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if leap == 1:
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print '%-4d' % m
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h += 1
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if h % 10 == 0:
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print ''
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leap = 1
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print 'The total is %d' % h
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'''
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【程序13】
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题目:打印出所有的“水仙花数”,所谓“水仙花数”是指一个三位数,其各位数字立方和等于该数
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本身。例如:153是一个“水仙花数”,因为153=1的三次方+5的三次方+3的三次方。
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1.程序分析:利用for循环控制100-999个数,每个数分解出个位,十位,百位。
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2.程序源代码:
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'''
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for n in range(100,1001):
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i = n / 100
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j = n / 10 % 10
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k = n % 10
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if i * 100 + j * 10 + k == i + j ** 2 + k ** 3:
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print "%-5d" % n
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'''
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【程序14】
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题目:将一个正整数分解质因数。例如:输入90,打印出90=2*3*3*5。
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程序分析:对n进行分解质因数,应先找到一个最小的质数k,然后按下述步骤完成:
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(1)如果这个质数恰等于n,则说明分解质因数的过程已经结束,打印出即可。
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(2)如果n<>k,但n能被k整除,则应打印出k的值,并用n除以k的商,作为新的正整数你n,
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重复执行第一步。
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(3)如果n不能被k整除,则用k+1作为k的值,重复执行第一步。
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2.程序源代码:
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'''
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from sys import stdout
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n = int(raw_input("input number:\n"))
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print "n = %d" % n
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for i in range(2,n + 1):
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while n != i:
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if n % i == 0:
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stdout.write(str(i))
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stdout.write("*")
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n = n / i
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else:
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break
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print "%d" % n
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@ -0,0 +1,17 @@
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'''
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【程序15】
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题目:利用条件运算符的嵌套来完成此题:学习成绩>=90分的同学用A表示,60-89分之间的用B表示,
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60分以下的用C表示。
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1.程序分析:(a>b)?a:b这是条件运算符的基本例子。
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2.程序源代码:
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不支持这个运算符
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'''
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score = int(raw_input('input score:\n'))
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if score >= 90:
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grade = 'A'
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elif score >= 60:
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grade = 'B'
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else:
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grade = 'C'
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print '%d belongs to %s' % (score,grade)
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@ -0,0 +1,23 @@
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'''
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【程序17】
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题目:输入一行字符,分别统计出其中英文字母、空格、数字和其它字符的个数。
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1.程序分析:利用while语句,条件为输入的字符不为'\n'.
|
||||
|
||||
2.程序源代码:
|
||||
'''
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import string
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s = raw_input('input a string:\n')
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letters = 0
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||||
space = 0
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||||
digit = 0
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others = 0
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for c in s:
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||||
if c.isalpha():
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||||
letters += 1
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||||
elif c.isspace():
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||||
space += 1
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elif c.isdigit():
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digit += 1
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else:
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others += 1
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||||
print 'char = %d,space = %d,digit = %d,others = %d' % (letters,space,digit,others)
|
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@ -0,0 +1,18 @@
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|||
'''
|
||||
题目:求s=a+aa+aaa+aaaa+aa...a的值,其中a是一个数字。例如2+22+222+2222+22222(此时
|
||||
共有5个数相加),几个数相加有键盘控制。
|
||||
1.程序分析:关键是计算出每一项的值。
|
||||
2.程序源代码:
|
||||
'''
|
||||
Tn = 0
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||||
Sn = []
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||||
n = int(raw_input('n = :\n'))
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||||
a = int(raw_input('a = :\n'))
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for count in range(n):
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Tn = Tn + a
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a = a * 10
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Sn.append(Tn)
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print Tn
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||||
|
||||
Sn = reduce(lambda x,y : x + y,Sn)
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print Sn
|
|
@ -0,0 +1,25 @@
|
|||
'''
|
||||
【程序19】
|
||||
题目:一个数如果恰好等于它的因子之和,这个数就称为“完数”。例如6=1+2+3.编程
|
||||
找出1000以内的所有完数。
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||||
1. 程序分析:请参照程序<--上页程序14.
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||||
2.程序源代码:
|
||||
'''
|
||||
from sys import stdout
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||||
for j in range(2,1001):
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||||
k = []
|
||||
n = -1
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||||
s = j
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||||
for i in range(1,j):
|
||||
if j % i == 0:
|
||||
n += 1
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||||
s -= i
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||||
k.append(i)
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||||
|
||||
if s == 0:
|
||||
print j
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||||
for i in range(n):
|
||||
stdout.write(k[i])
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||||
stdout.write(' ')
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||||
print k[n]
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||||
|
|
@ -0,0 +1,16 @@
|
|||
'''
|
||||
【程序20】
|
||||
题目:一球从100米高度自由落下,每次落地后反跳回原高度的一半;再落下,求它在
|
||||
第10次落地时,共经过多少米?第10次反弹多高?
|
||||
1.程序分析:见下面注释
|
||||
2.程序源代码:
|
||||
'''
|
||||
Sn = 100.0
|
||||
Hn = Sn / 2
|
||||
|
||||
for n in range(2,11):
|
||||
Sn += 2 * Hn
|
||||
Hn /= 2
|
||||
|
||||
print 'Total of road is %f' % Sn
|
||||
print 'The tenth is %f meter' % Hn
|
|
@ -0,0 +1,13 @@
|
|||
'''
|
||||
【程序21】
|
||||
题目:猴子吃桃问题:猴子第一天摘下若干个桃子,当即吃了一半,还不瘾,又多吃了一个
|
||||
第二天早上又将剩下的桃子吃掉一半,又多吃了一个。以后每天早上都吃了前一天剩下
|
||||
的一半零一个。到第10天早上想再吃时,见只剩下一个桃子了。求第一天共摘了多少。
|
||||
1.程序分析:采取逆向思维的方法,从后往前推断。
|
||||
2.程序源代码:
|
||||
'''
|
||||
x2 = 1
|
||||
for day in range(9,0,-1):
|
||||
x1 = (x2 + 1) * 2
|
||||
x2 = x1
|
||||
print x1
|
|
@ -0,0 +1,16 @@
|
|||
'''
|
||||
【程序22】
|
||||
题目:两个乒乓球队进行比赛,各出三人。甲队为a,b,c三人,乙队为x,y,z三人。已抽签决定
|
||||
比赛名单。有人向队员打听比赛的名单。a说他不和x比,c说他不和x,z比,请编程序找出
|
||||
三队赛手的名单。
|
||||
1.程序分析:判断素数的方法:用一个数分别去除2到sqrt(这个数),如果能被整除,
|
||||
则表明此数不是素数,反之是素数。
|
||||
2.程序源代码:
|
||||
'''
|
||||
for i in range(ord('x'),ord('z') + 1):
|
||||
for j in range(ord('x'),ord('z') + 1):
|
||||
if i != j:
|
||||
for k in range(ord('x'),ord('z') + 1):
|
||||
if (i != k) and (j != k):
|
||||
if (i != ord('x')) and (k != ord('x')) and (k != ord('z')):
|
||||
print 'order is a -- %s\t b -- %s\tc--%s' % (chr(i),chr(j),chr(k))
|
|
@ -0,0 +1,29 @@
|
|||
'''
|
||||
【程序23】
|
||||
题目:打印出如下图案(菱形)
|
||||
|
||||
*
|
||||
***
|
||||
*****
|
||||
*******
|
||||
*****
|
||||
***
|
||||
*
|
||||
1.程序分析:先把图形分成两部分来看待,前四行一个规律,后三行一个规律,利用双重
|
||||
for循环,第一层控制行,第二层控制列。
|
||||
2.程序源代码:
|
||||
'''
|
||||
from sys import stdout
|
||||
for i in range(4):
|
||||
for j in range(2 - i + 1):
|
||||
stdout.write(' ')
|
||||
for k in range(2 * i + 1):
|
||||
stdout.write('*')
|
||||
print
|
||||
|
||||
for i in range(3):
|
||||
for j in range(i + 1):
|
||||
stdout.write(' ')
|
||||
for k in range(4 - 2 * i + 1):
|
||||
stdout.write('*')
|
||||
print
|
|
@ -0,0 +1,34 @@
|
|||
'''
|
||||
【程序24】
|
||||
题目:有一分数序列:2/1,3/2,5/3,8/5,13/8,21/13...求出这个数列的前20项之和。
|
||||
1.程序分析:请抓住分子与分母的变化规律。
|
||||
2.程序源代码:
|
||||
'''
|
||||
#方法一
|
||||
a = 2.0
|
||||
b = 1.0
|
||||
s = 0
|
||||
for n in range(1,21):
|
||||
s += a / b
|
||||
t = a
|
||||
a = a + b
|
||||
b = t
|
||||
print s
|
||||
#方法二
|
||||
s = 0.0
|
||||
for n in range(1,21):
|
||||
s += a / b
|
||||
b,a = a , a + b
|
||||
print s
|
||||
|
||||
s = 0.0
|
||||
for n in range(1,21):
|
||||
s += a / b
|
||||
b,a = a , a + b
|
||||
print s
|
||||
#方法三
|
||||
l = []
|
||||
for n in range(1,21):
|
||||
b,a = a,a + b
|
||||
l.append(a / b)
|
||||
print reduce(lambda x,y: x + y,l)
|
|
@ -0,0 +1,25 @@
|
|||
'''
|
||||
【程序25】
|
||||
题目:求1+2!+3!+...+20!的和
|
||||
1.程序分析:此程序只是把累加变成了累乘。
|
||||
2.程序源代码:
|
||||
'''
|
||||
#方法一
|
||||
n = 0
|
||||
s = 0
|
||||
t = 1
|
||||
for n in range(1,21):
|
||||
t *= n
|
||||
s += t
|
||||
print '1! + 2! + 3! + ... + 20! = %d' % s
|
||||
|
||||
#方法二
|
||||
s = 0
|
||||
l = range(1,21)
|
||||
def op(x):
|
||||
r = 1
|
||||
for i in range(1,x + 1):
|
||||
r *= i
|
||||
return r
|
||||
s = sum(map(op,l))
|
||||
print '1! + 2! + 3! + ... + 20! = %d' % s
|
|
@ -0,0 +1,16 @@
|
|||
'''
|
||||
【程序26】
|
||||
题目:利用递归方法求5!。
|
||||
1.程序分析:递归公式:fn=fn_1*4!
|
||||
2.程序源代码:
|
||||
'''
|
||||
def fact(j):
|
||||
sum = 0
|
||||
if j == 0:
|
||||
sum = 1
|
||||
else:
|
||||
sum = j * fact(j - 1)
|
||||
return sum
|
||||
|
||||
for i in range(5):
|
||||
print '%d! = %d' % (i,fact(i))
|
|
@ -0,0 +1,20 @@
|
|||
'''
|
||||
【程序27】
|
||||
题目:利用递归函数调用方式,将所输入的5个字符,以相反顺序打印出来。
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
def palin(n):
|
||||
next = 0
|
||||
if n <= 1:
|
||||
next = input()
|
||||
print
|
||||
print next
|
||||
else:
|
||||
next = input()
|
||||
palin(n - 1)
|
||||
print next
|
||||
|
||||
i = 5
|
||||
palin(i)
|
||||
print
|
|
@ -0,0 +1,13 @@
|
|||
'''
|
||||
【程序28】
|
||||
题目:有5个人坐在一起,问第五个人多少岁?他说比第4个人大2岁。问第4个人岁数,他说比第
|
||||
3个人大2岁。问第三个人,又说比第2人大两岁。问第2个人,说比第一个人大两岁。最后
|
||||
问第一个人,他说是10岁。请问第五个人多大?
|
||||
1.程序分析:利用递归的方法,递归分为回推和递推两个阶段。要想知道第五个人岁数,需知道
|
||||
第四人的岁数,依次类推,推到第一人(10岁),再往回推。
|
||||
'''
|
||||
def age(n):
|
||||
if n == 1: c = 10
|
||||
else: c = age(n - 1) + 2
|
||||
return c
|
||||
print age(5)
|
|
@ -0,0 +1,23 @@
|
|||
'''
|
||||
【程序29】
|
||||
题目:给一个不多于5位的正整数,要求:一、求它是几位数,二、逆序打印出各位数字。
|
||||
1. 程序分析:学会分解出每一位数,如下解释:(这里是一种简单的算法,师专数002班赵鑫提供)
|
||||
2.程序源代码:
|
||||
'''
|
||||
x = int(raw_input("input a number:\n"))
|
||||
a = x / 10000
|
||||
b = x % 10000 / 1000
|
||||
c = x % 1000 / 100
|
||||
d = x % 100 / 10
|
||||
e = x % 10
|
||||
|
||||
if a != 0:
|
||||
print "there are 5 ",e,d,c,b,a
|
||||
elif b != 0:
|
||||
print "there are 4 ",d,c,b,a
|
||||
elif c != 0:
|
||||
print "there are 3 ",e,d,c
|
||||
elif d != 0:
|
||||
print "there are 2 ",e,d
|
||||
else:
|
||||
print "there are 1",e
|
|
@ -0,0 +1,12 @@
|
|||
'''
|
||||
题目:一个5位数,判断它是不是回文数。即12321是回文数,个位与万位相同,十位与千位相同。
|
||||
1.程序分析:同29例
|
||||
2.程序源代码:
|
||||
'''
|
||||
x = int(raw_input("input a number:\n"))
|
||||
x = str(x)
|
||||
for i in range(len(x)/2):
|
||||
if x[i] != x[-i - 1]:
|
||||
print 'this number is not a huiwen'
|
||||
break
|
||||
print 'this number is a huiwen'
|
|
@ -0,0 +1,46 @@
|
|||
'''
|
||||
程序31】
|
||||
题目:请输入星期几的第一个字母来判断一下是星期几,如果第一个字母一样,则继续
|
||||
判断第二个字母。
|
||||
1.程序分析:用情况语句比较好,如果第一个字母一样,则判断用情况语句或if语句判断第二个字母。
|
||||
2.程序源代码:
|
||||
'''
|
||||
from sys import stdin
|
||||
letter = stdin.read(1)
|
||||
stdin.flush()
|
||||
while letter != 'Y':
|
||||
if letter == 'S':
|
||||
print 'please input second letter'
|
||||
letter = stdin.read(1)
|
||||
stdin.flush()
|
||||
if letter == 'a':
|
||||
print 'Saturday'
|
||||
elif letter == 'u':
|
||||
print 'Sunday'
|
||||
else:
|
||||
print 'data error'
|
||||
break
|
||||
elif letter == 'F':
|
||||
print 'Friday'
|
||||
break
|
||||
elif letter == 'M':
|
||||
print 'Monday'
|
||||
#break
|
||||
elif letter == 'T':
|
||||
print 'please input second letter'
|
||||
letter = stdin.read(1)
|
||||
stdin.flush()
|
||||
if letter == 'u':
|
||||
print 'Tuesday'
|
||||
elif letter == 'h':
|
||||
print 'Thursday'
|
||||
else:
|
||||
print 'data error'
|
||||
break
|
||||
elif letter == 'W':
|
||||
print 'Wednesday'
|
||||
else:
|
||||
print 'data error'
|
||||
letter = stdin.read(1)
|
||||
stdin.flush()
|
||||
|
|
@ -0,0 +1,7 @@
|
|||
'''
|
||||
【程序32】
|
||||
题目:Press any key to change color, do you want to try it. Please hurry up!
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
不知道写呢 :(,先空着吧
|
||||
'''
|
|
@ -0,0 +1,7 @@
|
|||
'''
|
||||
【程序33】
|
||||
题目:学习gotoxy()与clrscr()函数
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
不知道如何写 :( 先空着吧
|
||||
'''
|
|
@ -0,0 +1,14 @@
|
|||
'''
|
||||
【程序34】
|
||||
题目:练习函数调用
|
||||
1. 程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
def hello_world():
|
||||
print 'hello world'
|
||||
|
||||
def three_hellos():
|
||||
for i in range(3):
|
||||
hello_world()
|
||||
if __name__ == '__main__':
|
||||
three_hellos()
|
|
@ -0,0 +1,18 @@
|
|||
'''
|
||||
【程序35】
|
||||
题目:文本颜色设置
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
#include <conio.h>
|
||||
void main(void)
|
||||
{
|
||||
int color;
|
||||
for (color = 1; color < 16; color++)
|
||||
{
|
||||
textcolor(color);/*设置文本颜色*/
|
||||
cprintf("This is color %d\r\n", color);
|
||||
}
|
||||
textcolor(128 + 15);
|
||||
cprintf("This is blinking\r\n");
|
||||
}
|
||||
'''
|
|
@ -0,0 +1,21 @@
|
|||
'''
|
||||
【程序36】
|
||||
题目:求100之内的素数
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
from math import sqrt
|
||||
if __name__ == '__main__':
|
||||
N = 100
|
||||
a = range(0,N)
|
||||
for i in range(2,int(sqrt(N))):
|
||||
for j in range(i + 1,N):
|
||||
if (a[i] != 0) and (a[j] != 0):
|
||||
if a[j] % a[i] == 0:
|
||||
a[j] = 0
|
||||
print
|
||||
for i in range(2,N):
|
||||
if a[i] != 0:
|
||||
print "%5d" % a[i]
|
||||
if (i - 2) % 10 == 0:
|
||||
print
|
|
@ -0,0 +1,29 @@
|
|||
'''
|
||||
【程序37】
|
||||
题目:对10个数进行排序
|
||||
1.程序分析:可以利用选择法,即从后9个比较过程中,选择一个最小的与第一个元素交换,
|
||||
下次类推,即用第二个元素与后8个进行比较,并进行交换。
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == "__main__":
|
||||
N = 10
|
||||
# input data
|
||||
print 'please input ten num:\n'
|
||||
l = []
|
||||
for i in range(N):
|
||||
l.append(int(raw_input('input a number:\n')))
|
||||
print
|
||||
for i in range(N):
|
||||
print l[i]
|
||||
print
|
||||
|
||||
# sort ten num
|
||||
for i in range(N - 1):
|
||||
min = i
|
||||
for j in range(i + 1,N):
|
||||
if l[min] > l[j]:min = j
|
||||
l[i],l[min] = l[min],l[i]
|
||||
print 'after sorted'
|
||||
for i in range(N):
|
||||
print l[i]
|
||||
|
|
@ -0,0 +1,17 @@
|
|||
'''
|
||||
【程序38】
|
||||
题目:求一个3*3矩阵对角线元素之和
|
||||
1.程序分析:利用双重for循环控制输入二维数组,再将a[i][i]累加后输出。
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
a = []
|
||||
sum = 0.0
|
||||
for i in range(3):
|
||||
a.append([])
|
||||
for j in range(3):
|
||||
a[i].append(float(raw_input("input num:\n")))
|
||||
for i in range(3):
|
||||
sum += a[i][i]
|
||||
print sum
|
||||
|
|
@ -0,0 +1,40 @@
|
|||
'''
|
||||
【程序39】
|
||||
题目:有一个已经排好序的数组。现输入一个数,要求按原来的规律将它插入数组中。
|
||||
1. 程序分析:首先判断此数是否大于最后一个数,然后再考虑插入中间的数的情况,插入后
|
||||
此元素之后的数,依次后移一个位置。
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
# 方法一
|
||||
a = [1,4,6,9,13,16,19,28,40,100,0]
|
||||
print 'original list is:'
|
||||
for i in range(len(a)):
|
||||
print a[i]
|
||||
number = int(raw_input("insert a new number:\n"))
|
||||
end = a[9]
|
||||
if number > end:
|
||||
a[10] = number
|
||||
else:
|
||||
for i in range(10):
|
||||
if a[i] > number:
|
||||
temp1 = a[i]
|
||||
a[i] = number
|
||||
for j in range(i + 1,11):
|
||||
temp2 = a[j]
|
||||
a[j] = temp1
|
||||
temp1 = temp2
|
||||
break
|
||||
for i in range(11):
|
||||
print a[i]
|
||||
# 方法二
|
||||
# insrt another number
|
||||
number = int(raw_input('input a number:\n'))
|
||||
if number > a[len(a) - 1]:
|
||||
a.append(number)
|
||||
else:
|
||||
for i in range(len(a)):
|
||||
if a[i] > number:
|
||||
a.insert(i,number)
|
||||
print a
|
||||
|
|
@ -0,0 +1,13 @@
|
|||
'''
|
||||
【程序40】
|
||||
题目:将一个数组逆序输出。
|
||||
1.程序分析:用第一个与最后一个交换。
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
a = [9,6,5,4,1]
|
||||
N = len(a)
|
||||
print a
|
||||
for i in range(len(a) / 2):
|
||||
a[i],a[N - i - 1] = a[N - i - 1],a[i]
|
||||
print a
|
|
@ -0,0 +1,27 @@
|
|||
'''
|
||||
【程序41】
|
||||
题目:学习static定义静态变量的用法
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
# python没有这个功能了,只能这样了:)
|
||||
def varfunc():
|
||||
var = 0
|
||||
print 'var = %d' % var
|
||||
var += 1
|
||||
if __name__ == '__main__':
|
||||
for i in range(3):
|
||||
varfunc()
|
||||
|
||||
# attribut of class
|
||||
# 作为类的一个属性吧
|
||||
class Static:
|
||||
StaticVar = 5
|
||||
def varfunc(self):
|
||||
self.StaticVar += 1
|
||||
print self.StaticVar
|
||||
|
||||
print Static.StaticVar
|
||||
a = Static()
|
||||
for i in range(3):
|
||||
a.varfunc()
|
|
@ -0,0 +1,16 @@
|
|||
'''
|
||||
题目:学习使用auto定义变量的用法
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
没有auto关键字,使用变量作用域来举例吧
|
||||
'''
|
||||
num = 2
|
||||
def autofunc():
|
||||
num = 1
|
||||
print 'internal block num = %d' % num
|
||||
num += 1
|
||||
for i in range(3):
|
||||
print 'The num = %d' % num
|
||||
num += 1
|
||||
autofunc()
|
||||
|
|
@ -0,0 +1,20 @@
|
|||
'''
|
||||
【程序43】
|
||||
题目:学习使用static的另一用法。
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
有一个static变量的用法,python是没有,演示一个python作用域使用方法
|
||||
'''
|
||||
class Num:
|
||||
nNum = 1
|
||||
def inc(self):
|
||||
self.nNum += 1
|
||||
print 'nNum = %d' % self.nNum
|
||||
|
||||
if __name__ == '__main__':
|
||||
nNum = 2
|
||||
inst = Num()
|
||||
for i in range(3):
|
||||
nNum += 1
|
||||
print 'The num = %d' % nNum
|
||||
inst.inc()
|
|
@ -0,0 +1,11 @@
|
|||
'''
|
||||
【程序44】
|
||||
题目:学习使用external的用法。
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
external.py代码:
|
||||
|
||||
'''
|
||||
import external
|
||||
if __name__ == '__main__':
|
||||
print external.add(10,20)
|
|
@ -0,0 +1,11 @@
|
|||
'''
|
||||
【程序45】
|
||||
题目:学习使用register定义变量的方法。
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
没有register关键字,用整型变量代替
|
||||
'''
|
||||
tmp = 0
|
||||
for i in range(1,101):
|
||||
tmp += i
|
||||
print 'The sum is %d' % tmp
|
|
@ -0,0 +1,20 @@
|
|||
'''
|
||||
【程序46】
|
||||
题目:宏#define命令练习(1)
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
没有C语言的宏,就这么写了
|
||||
'''
|
||||
TRUE = 1
|
||||
FALSE = 0
|
||||
def SQ(x):
|
||||
return x * x
|
||||
print 'Program will stop if input value less than 50.'
|
||||
again = 1
|
||||
while again:
|
||||
num = int(raw_input('Please input number'))
|
||||
print 'The square for this number is %d' % (SQ(num))
|
||||
if num >= 50:
|
||||
again = TRUE
|
||||
else:
|
||||
again = FALSE
|
|
@ -0,0 +1,23 @@
|
|||
'''
|
||||
题目:宏#define命令练习(2)
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
#include "stdio.h"
|
||||
#define exchange(a,b) { \ /*宏定义中允许包含两道衣裳命令的情形,此时必须在最右边加上"\"*/
|
||||
int t;\
|
||||
t=a;\
|
||||
a=b;\
|
||||
b=t;\
|
||||
}'
|
||||
这个宏定义python不支持
|
||||
'''
|
||||
def exchange(a,b):
|
||||
a,b = b,a
|
||||
return (a,b)
|
||||
|
||||
if __name__ == '__main__':
|
||||
x = 10
|
||||
y = 20
|
||||
print 'x = %d,y = %d' % (x,y)
|
||||
x,y = exchange(x,y)
|
||||
print 'x = %d,y = %d' % (x,y)
|
|
@ -0,0 +1,36 @@
|
|||
'''
|
||||
【程序48】
|
||||
题目:宏#define命令练习(3)
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
#define LAG >
|
||||
#define SMA <
|
||||
#define EQ ==
|
||||
#include "stdio.h"
|
||||
void main()
|
||||
{
|
||||
int i=10;
|
||||
int j=20;
|
||||
if(i LAG j)
|
||||
printf("\40: %d larger than %d \n",i,j);
|
||||
else if(i EQ j)
|
||||
printf("\40: %d equal to %d \n",i,j);
|
||||
else if(i SMA j)
|
||||
printf("\40:%d smaller than %d \n",i,j);
|
||||
else
|
||||
printf("\40: No such value.\n");
|
||||
}
|
||||
不知道如何用python实现类似的功能
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
i = 10
|
||||
j = 20
|
||||
if i > j:
|
||||
print '%d larger than %d' % (i,j)
|
||||
elif i == j:
|
||||
print '%d equal to %d' % (i,j)
|
||||
elif i < j:
|
||||
print '%d smaller than %d' % (i,j)
|
||||
else:
|
||||
print 'No such value'
|
||||
|
|
@ -0,0 +1,45 @@
|
|||
'''
|
||||
【程序49】
|
||||
题目:#if #ifdef和#ifndef的综合应用。
|
||||
1. 程序分析:
|
||||
2.程序源代码:
|
||||
#include "stdio.h"
|
||||
#define MAX
|
||||
#define MAXIMUM(x,y) (x>y)?x:y
|
||||
#define MINIMUM(x,y) (x>y)?y:x
|
||||
void main()
|
||||
{
|
||||
int a=10,b=20;
|
||||
#ifdef MAX
|
||||
printf("\40: The larger one is %d\n",MAXIMUM(a,b));
|
||||
#else
|
||||
printf("\40: The lower one is %d\n",MINIMUM(a,b));
|
||||
#endif
|
||||
#ifndef MIN
|
||||
printf("\40: The lower one is %d\n",MINIMUM(a,b));
|
||||
#else
|
||||
printf("\40: The larger one is %d\n",MAXIMUM(a,b));
|
||||
#endif
|
||||
#undef MAX
|
||||
#ifdef MAX
|
||||
printf("\40: The larger one is %d\n",MAXIMUM(a,b));
|
||||
#else
|
||||
printf("\40: The lower one is %d\n",MINIMUM(a,b));
|
||||
#endif
|
||||
#define MIN
|
||||
#ifndef MIN
|
||||
printf("\40: The lower one is %d\n",MINIMUM(a,b));
|
||||
#else
|
||||
printf("\40: The larger one is %d\n",MAXIMUM(a,b));
|
||||
#endif
|
||||
}
|
||||
这个还是预处理的用法,python不支持这样的机制,演示lambda的使用。
|
||||
'''
|
||||
MAXIMUM = lambda x,y : (x > y) * x + (x < y) * y
|
||||
MINIMUM = lambda x,y : (x > y) * y + (x < y) * x
|
||||
|
||||
if __name__ == '__main__':
|
||||
a = 10
|
||||
b = 20
|
||||
print 'The largar one is %d' % MAXIMUM(a,b)
|
||||
print 'The lower one is %d' % MINIMUM(a,b)
|
|
@ -0,0 +1,12 @@
|
|||
'''
|
||||
【程序51】
|
||||
题目:学习使用按位与 & 。
|
||||
1.程序分析:0&0=0; 0&1=0; 1&0=0; 1&1=1
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
a = 077
|
||||
b = a & 3
|
||||
print 'a & b = %d' % b
|
||||
b &= 7
|
||||
print 'a & b = %d' % b
|
|
@ -0,0 +1,12 @@
|
|||
'''
|
||||
题目:学习使用按位或 | 。
|
||||
1.程序分析:0|0=0; 0|1=1; 1|0=1; 1|1=1
|
||||
2.程序源代码:
|
||||
'''
|
||||
|
||||
if __name__ == '__main__':
|
||||
a = 077
|
||||
b = a | 3
|
||||
print 'a | b is %d' % b
|
||||
b |= 7
|
||||
print 'a | b is %d' % b
|
|
@ -0,0 +1,12 @@
|
|||
'''
|
||||
【程序53】
|
||||
题目:学习使用按位异或 ^ 。
|
||||
1.程序分析:0^0=0; 0^1=1; 1^0=1; 1^1=0
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
a = 077
|
||||
b = a ^ 3
|
||||
print 'The a ^ 3 = %d' % b
|
||||
b ^= 7
|
||||
print 'The a ^ b = %d' % b
|
|
@ -0,0 +1,14 @@
|
|||
'''
|
||||
【程序54】
|
||||
题目:取一个整数a从右端开始的4~7位。
|
||||
程序分析:可以这样考虑:
|
||||
(1)先使a右移4位。
|
||||
(2)设置一个低4位全为1,其余全为0的数。可用~(~0<<4)
|
||||
(3)将上面二者进行&运算。
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
a = int(raw_input('input a number:\n'))
|
||||
b = a >> 4
|
||||
c = ~(~0 << 4)
|
||||
d = b & c
|
||||
print '%o\t%o' %(a,d)
|
|
@ -0,0 +1,13 @@
|
|||
'''
|
||||
【程序55】
|
||||
题目:学习使用按位取反~。
|
||||
1.程序分析:~0=1; ~1=0;
|
||||
2.程序源代码:
|
||||
如何查看复数的16进制数
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
a = 234
|
||||
b = ~a
|
||||
print 'The a\'s 1 complement is %d' % b
|
||||
a = ~a
|
||||
print 'The a\'s 2 complement is %d' % a
|
|
@ -0,0 +1,36 @@
|
|||
'''
|
||||
【程序56】
|
||||
题目:画图,学用circle画圆形。
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
#include "graphics.h"
|
||||
main()
|
||||
{
|
||||
int driver,mode,i;
|
||||
float j=1,k=1;
|
||||
driver=VGA;mode=VGAHI;
|
||||
initgraph(&driver,&mode,"");
|
||||
setbkcolor(YELLOW);
|
||||
for(i=0;i<=25;i++)
|
||||
{
|
||||
setcolor(8);
|
||||
circle(310,250,k);
|
||||
k=k+j;
|
||||
j=j+0.3;
|
||||
}
|
||||
}
|
||||
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
from Tkinter import *
|
||||
|
||||
canvas = Canvas(width=800, height=600, bg='yellow')
|
||||
canvas.pack(expand=YES, fill=BOTH)
|
||||
k = 1
|
||||
j = 1
|
||||
for i in range(0,26):
|
||||
canvas.create_oval(310 - k,250 - k,310 + k,250 + k, width=1)
|
||||
k += j
|
||||
j += 0.3
|
||||
|
||||
mainloop()
|
|
@ -0,0 +1,32 @@
|
|||
'''
|
||||
【程序57】
|
||||
题目:画图,学用line画直线。
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
from Tkinter import *
|
||||
|
||||
canvas = Canvas(width=300, height=300, bg='green')
|
||||
canvas.pack(expand=YES, fill=BOTH)
|
||||
x0 = 263
|
||||
y0 = 263
|
||||
y1 = 275
|
||||
x1 = 275
|
||||
for i in range(19):
|
||||
canvas.create_line(x0,y0,x0,y1, width=1, fill='red')
|
||||
x0 = x0 - 5
|
||||
y0 = y0 - 5
|
||||
x1 = x1 + 5
|
||||
y1 = y1 + 5
|
||||
|
||||
x0 = 263
|
||||
y1 = 275
|
||||
y0 = 263
|
||||
for i in range(21):
|
||||
canvas.create_line(x0,y0,x0,y1,fill = 'red')
|
||||
x0 += 5
|
||||
y0 += 5
|
||||
y1 += 5
|
||||
|
||||
mainloop()
|
|
@ -0,0 +1,24 @@
|
|||
'''
|
||||
【程序58】
|
||||
题目:画图,学用rectangle画方形。
|
||||
1.程序分析:利用for循环控制100-999个数,每个数分解出个位,十位,百位。
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
from Tkinter import *
|
||||
root = Tk()
|
||||
root.title('Canvas')
|
||||
canvas = Canvas(root,width = 400,height = 400,bg = 'yellow')
|
||||
x0 = 263
|
||||
y0 = 263
|
||||
y1 = 275
|
||||
x1 = 275
|
||||
for i in range(19):
|
||||
canvas.create_rectangle(x0,y0,x1,y1)
|
||||
x0 -= 5
|
||||
y0 -= 5
|
||||
x1 += 5
|
||||
y1 += 5
|
||||
|
||||
canvas.pack()
|
||||
root.mainloop()
|
|
@ -0,0 +1,36 @@
|
|||
'''
|
||||
题目:画图,综合例子。
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
from Tkinter import *
|
||||
canvas = Canvas(width = 300,height = 300,bg = 'green')
|
||||
canvas.pack(expand = YES,fill = BOTH)
|
||||
x0 = 150
|
||||
y0 = 100
|
||||
canvas.create_oval(x0 - 10,y0 - 10,x0 + 10,y0 + 10)
|
||||
canvas.create_oval(x0 - 20,y0 - 20,x0 + 20,y0 + 20)
|
||||
canvas.create_oval(x0 - 50,y0 - 50,x0 + 50,y0 + 50)
|
||||
import math
|
||||
B = 0.809
|
||||
for i in range(16):
|
||||
a = 2 * math.pi / 16 * i
|
||||
x = math.ceil(x0 + 48 * math.cos(a))
|
||||
y = math.ceil(y0 + 48 * math.sin(a) * B)
|
||||
canvas.create_line(x0,y0,x,y,fill = 'red')
|
||||
canvas.create_oval(x0 - 60,y0 - 60,x0 + 60,y0 + 60)
|
||||
|
||||
|
||||
for k in range(501):
|
||||
for i in range(17):
|
||||
a = (2 * math.pi / 16) * i + (2 * math.pi / 180) * k
|
||||
x = math.ceil(x0 + 48 * math.cos(a))
|
||||
y = math.ceil(y0 + 48 + math.sin(a) * B)
|
||||
canvas.create_line(x0,y0,x,y,fill = 'red')
|
||||
for j in range(51):
|
||||
a = (2 * math.pi / 16) * i + (2* math.pi / 180) * k - 1
|
||||
x = math.ceil(x0 + 48 * math.cos(a))
|
||||
y = math.ceil(y0 + 48 * math.sin(a) * B)
|
||||
canvas.create_line(x0,y0,x,y,fill = 'red')
|
||||
mainloop()
|
|
@ -0,0 +1,49 @@
|
|||
'''
|
||||
【程序60】
|
||||
题目:画图,综合例子。
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
键盘不知道如何响应,先不写这个
|
||||
#include "graphics.h"
|
||||
#define LEFT 0
|
||||
#define TOP 0
|
||||
#define RIGHT 639
|
||||
#define BOTTOM 479
|
||||
#define LINES 400
|
||||
#define MAXCOLOR 15
|
||||
main()
|
||||
{
|
||||
int driver,mode,error;
|
||||
int x1,y1;
|
||||
int x2,y2;
|
||||
int dx1,dy1,dx2,dy2,i=1;
|
||||
int count=0;
|
||||
int color=0;
|
||||
driver=VGA;
|
||||
mode=VGAHI;
|
||||
initgraph(&driver,&mode,"");
|
||||
x1=x2=y1=y2=10;
|
||||
dx1=dy1=2;
|
||||
dx2=dy2=3;
|
||||
while(!kbhit())
|
||||
{
|
||||
line(x1,y1,x2,y2);
|
||||
x1+=dx1;y1+=dy1;
|
||||
x2+=dx2;y2+dy2;
|
||||
if(x1<=LEFT||x1>=RIGHT)
|
||||
dx1=-dx1;
|
||||
if(y1<=TOP||y1>=BOTTOM)
|
||||
dy1=-dy1;
|
||||
if(x2<=LEFT||x2>=RIGHT)
|
||||
dx2=-dx2;
|
||||
if(y2<=TOP||y2>=BOTTOM)
|
||||
dy2=-dy2;
|
||||
if(++count>LINES)
|
||||
{
|
||||
setcolor(color);
|
||||
color=(color>=MAXCOLOR)?0:++color;
|
||||
}
|
||||
}
|
||||
closegraph();
|
||||
}
|
||||
''
|
|
@ -0,0 +1,25 @@
|
|||
'''
|
||||
【程序61】
|
||||
题目:打印出杨辉三角形(要求打印出10行如下图)
|
||||
1.程序分析:
|
||||
'''
|
||||
|
||||
if __name__ == '__main__':
|
||||
a = []
|
||||
for i in range(10):
|
||||
a.append([])
|
||||
for j in range(10):
|
||||
a[i].append(0)
|
||||
for i in range(10):
|
||||
a[i][0] = 1
|
||||
a[i][i] = 1
|
||||
for i in range(2,10):
|
||||
for j in range(1,i):
|
||||
a[i][j] = a[i - 1][j-1] + a[i - 1][j]
|
||||
from sys import stdout
|
||||
for i in range(10):
|
||||
for j in range(i + 1):
|
||||
stdout.write(a[i][j])
|
||||
stdout.write(' ')
|
||||
print
|
||||
|
|
@ -0,0 +1,20 @@
|
|||
'''
|
||||
【程序62】
|
||||
题目:学习putpixel画点。
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
#include "stdio.h"
|
||||
#include "graphics.h"
|
||||
main()
|
||||
{
|
||||
int i,j,driver=VGA,mode=VGAHI;
|
||||
initgraph(&driver,&mode,"");
|
||||
setbkcolor(YELLOW);
|
||||
for(i=50;i<=230;i+=20)
|
||||
for(j=50;j<=230;j++)
|
||||
putpixel(i,j,1);
|
||||
for(j=50;j<=230;j+=20)
|
||||
for(i=50;i<=230;i++)
|
||||
putpixel(i,j,1);
|
||||
}
|
||||
'''
|
|
@ -0,0 +1,19 @@
|
|||
'''
|
||||
题目:画椭圆ellipse
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
from Tkinter import *
|
||||
x = 360
|
||||
y = 160
|
||||
top = y - 30
|
||||
bottom = y - 30
|
||||
|
||||
canvas = Canvas(width = 400,height = 600,bg = 'white')
|
||||
for i in range(20):
|
||||
canvas.create_oval(250 - top,250 - bottom,250 + top,250 + bottom)
|
||||
top -= 5
|
||||
bottom += 5
|
||||
canvas.pack()
|
||||
mainloop()
|
|
@ -0,0 +1,23 @@
|
|||
'''
|
||||
题目:利用ellipse and rectangle 画图。
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
from Tkinter import *
|
||||
canvas = Canvas(width = 400,height = 600,bg = 'white')
|
||||
left = 20
|
||||
right = 50
|
||||
top = 50
|
||||
num = 15
|
||||
for i in range(num):
|
||||
canvas.create_oval(250 - right,250 - left,250 + right,250 + left)
|
||||
canvas.create_oval(250 - 20,250 - top,250 + 20,250 + top)
|
||||
canvas.create_rectangle(20 - 2 * i,20 - 2 * i,10 * (i + 2),10 * ( i + 2))
|
||||
right += 5
|
||||
left += 5
|
||||
top += 10
|
||||
|
||||
canvas.pack()
|
||||
mainloop()
|
||||
|
|
@ -0,0 +1,45 @@
|
|||
'''
|
||||
【程序65】
|
||||
题目:一个最优美的图案。
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
import math
|
||||
class PTS:
|
||||
def __init__(self):
|
||||
self.x = 0
|
||||
self.y = 0
|
||||
points = []
|
||||
|
||||
def LineToDemo():
|
||||
from Tkinter import *
|
||||
screenx = 400
|
||||
screeny = 400
|
||||
canvas = Canvas(width = screenx,height = screeny,bg = 'white')
|
||||
|
||||
AspectRatio = 0.85
|
||||
MAXPTS = 15
|
||||
h = screeny
|
||||
w = screenx
|
||||
xcenter = w / 2
|
||||
ycenter = h / 2
|
||||
radius = (h - 30) / (AspectRatio * 2) - 20
|
||||
step = 360 / MAXPTS
|
||||
angle = 0.0
|
||||
for i in range(MAXPTS):
|
||||
rads = angle * math.pi / 180.0
|
||||
p = PTS()
|
||||
p.x = xcenter + int(math.cos(rads) * radius)
|
||||
p.y = ycenter - int(math.sin(rads) * radius * AspectRatio)
|
||||
angle += step
|
||||
points.append(p)
|
||||
canvas.create_oval(xcenter - radius,ycenter - radius,
|
||||
xcenter + radius,ycenter + radius)
|
||||
for i in range(MAXPTS):
|
||||
for j in range(i,MAXPTS):
|
||||
canvas.create_line(points[i].x,points[i].y,points[j].x,points[j].y)
|
||||
|
||||
canvas.pack()
|
||||
mainloop()
|
||||
if __name__ == '__main__':
|
||||
LineToDemo()
|
|
@ -0,0 +1,19 @@
|
|||
'''
|
||||
【程序66】
|
||||
题目:输入3个数a,b,c,按大小顺序输出。
|
||||
1.程序分析:利用指针方法。
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
n1 = int(raw_input('n1 = :\n'))
|
||||
n2 = int(raw_input('n2 = :\n'))
|
||||
n3 = int(raw_input('n3 = :\n'))
|
||||
|
||||
def swap(p1,p2):
|
||||
return p2,p1
|
||||
|
||||
if n1 > n2 : n1,n2 = swap(n1,n2)
|
||||
if n1 > n3 : n1,n3 = swap(n1,n3)
|
||||
if n2 > n3 : n2,n3 = swap(n2,n3)
|
||||
|
||||
print n1,n2,n3
|
|
@ -0,0 +1,32 @@
|
|||
'''
|
||||
题目:输入数组,最大的与第一个元素交换,最小的与最后一个元素交换,输出数组。
|
||||
1.程序分析:谭浩强的书中答案有问题。
|
||||
2.程序源代码:
|
||||
'''
|
||||
def inp(numbers):
|
||||
for i in range(9):
|
||||
numbers.append(int(raw_input('input a number:\n')))
|
||||
numbers.append(int(raw_input('input a number:\n')))
|
||||
p = 0
|
||||
def max_min(array):
|
||||
max = min = 0
|
||||
for i in range(1,len(array) - 1):
|
||||
p = i
|
||||
if array[p] > array[max] : max = p
|
||||
elif array[p] < array[min] : min = p
|
||||
k = max
|
||||
l = min
|
||||
array[0],array[l] = array[l],array[0]
|
||||
array[9],array[k] = array[k],array[9]
|
||||
|
||||
def outp(numbers):
|
||||
for i in range(len(numbers)):
|
||||
print numbers[i]
|
||||
|
||||
if __name__ == '__main__':
|
||||
array = []
|
||||
inp(array)
|
||||
max_min(array)
|
||||
outp(array)
|
||||
|
||||
|
|
@ -0,0 +1,26 @@
|
|||
'''
|
||||
【程序68】
|
||||
题目:有n个整数,使其前面各数顺序向后移m个位置,最后m个数变成最前面的m个数
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
n = int(raw_input('the total number is:\n'))
|
||||
m = int(raw_input('back m:\n'))
|
||||
|
||||
def move(array,n,m):
|
||||
array_end = array[n - 1]
|
||||
for i in range(n - 1,-1,- 1):
|
||||
array[i] = array[i - 1]
|
||||
array[0] = array_end
|
||||
m -= 1
|
||||
if m > 0:move(array,n,m)
|
||||
|
||||
number = []
|
||||
for i in range(n):
|
||||
number.append(int(raw_input('input a number:\n')))
|
||||
print 'orignal number:',number
|
||||
|
||||
move(number,n,m)
|
||||
|
||||
print 'after moved:',number
|
|
@ -0,0 +1,30 @@
|
|||
'''
|
||||
【程序69】
|
||||
题目:有n个人围成一圈,顺序排号。从第一个人开始报数(从1到3报数),凡报到3的人退出
|
||||
圈子,问最后留下的是原来第几号的那位。
|
||||
1. 程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
nmax = 50
|
||||
n = int(raw_input('please input the total of numbers:'))
|
||||
num = []
|
||||
for i in range(n):
|
||||
num.append(i + 1)
|
||||
|
||||
i = 0
|
||||
k = 0
|
||||
m = 0
|
||||
|
||||
while m < n - 1:
|
||||
if num[i] != 0 : k += 1
|
||||
if k == 3:
|
||||
num[i] = 0
|
||||
k = 0
|
||||
m += 0
|
||||
i += 1
|
||||
if i == n : i = 0
|
||||
|
||||
i = 0
|
||||
while num[i] == 0: i += 1
|
||||
print num[i]
|
|
@ -0,0 +1,9 @@
|
|||
'''
|
||||
题目:写一个函数,求一个字符串的长度,在main函数中输入字符串,并输出其长度。
|
||||
1.程序分析:
|
||||
2.程序源代码
|
||||
就这样吧
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
s = raw_input('please input a string:\n')
|
||||
print 'the string has %d characters.' % len(s)
|
|
@ -0,0 +1,34 @@
|
|||
'''
|
||||
【程序71】
|
||||
题目:编写input()和output()函数输入,输出5个学生的数据记录。
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
使用list来模拟结构(不使用class)
|
||||
stu = [string,string,list]
|
||||
'''
|
||||
N = 3
|
||||
#stu
|
||||
# num : string
|
||||
# name : string
|
||||
# score[4]: list
|
||||
student = []
|
||||
for i in range(5):
|
||||
student.append(['','',[]])
|
||||
|
||||
def input_stu(stu):
|
||||
for i in range(N):
|
||||
stu[i][0] = raw_input('input student num:\n')
|
||||
stu[i][1] = raw_input('input student name:\n')
|
||||
for j in range(3):
|
||||
stu[i][2].append(int(raw_input('score:\n')))
|
||||
|
||||
def output_stu(stu):
|
||||
for i in range(N):
|
||||
print '%-6s%-10s' % ( stu[i][0],stu[i][1] )
|
||||
for j in range(3):
|
||||
print '%-8d' % stu[i][2][j]
|
||||
|
||||
if __name__ == '__main__':
|
||||
input_stu(student)
|
||||
print student
|
||||
output_stu(student)
|
|
@ -0,0 +1,12 @@
|
|||
'''
|
||||
【程序72】
|
||||
题目:创建一个链表。
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
ptr = []
|
||||
for i in range(5):
|
||||
num = int(raw_input('please input a number:\n'))
|
||||
ptr.append(num)
|
||||
print ptr
|
|
@ -0,0 +1,13 @@
|
|||
'''
|
||||
题目:反向输出一个链表。
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
ptr = []
|
||||
for i in range(5):
|
||||
num = int(raw_input('please input a number:\n'))
|
||||
ptr.append(num)
|
||||
print ptr
|
||||
ptr.reverse()
|
||||
print ptr
|
|
@ -0,0 +1,13 @@
|
|||
'''
|
||||
【程序74】
|
||||
题目:连接两个链表。
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
代码上好像只有,列表排序
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
arr1 = (3,12,8,9,11)
|
||||
ptr = list(arr1)
|
||||
print ptr
|
||||
ptr.sort()
|
||||
print ptr
|
|
@ -0,0 +1,14 @@
|
|||
'''
|
||||
【程序75】
|
||||
题目:放松一下,算一道简单的题目。
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
for i in range(5):
|
||||
n = 0
|
||||
if i != 1: n += 1
|
||||
if i == 3: n += 1
|
||||
if i == 4: n += 1
|
||||
if i != 4: n += 1
|
||||
if n == 3: print 64 + i
|
|
@ -0,0 +1,31 @@
|
|||
'''
|
||||
【程序76】
|
||||
题目:编写一个函数,输入n为偶数时,调用函数求1/2+1/4+...+1/n,当输入n为奇数时,调用函数
|
||||
1/1+1/3+...+1/n(利用指针函数)
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
def peven(n):
|
||||
i = 0
|
||||
s = 0.0
|
||||
for i in range(2,n + 1,2):
|
||||
s += 1.0 / i
|
||||
return s
|
||||
|
||||
def podd(n):
|
||||
s = 0.0
|
||||
for i in range(1, n + 1,2):
|
||||
s += 1 / i
|
||||
return s
|
||||
|
||||
def dcall(fp,n):
|
||||
s = fp(n)
|
||||
return s
|
||||
|
||||
if __name__ == '__main__':
|
||||
n = int(raw_input('input a number:\n'))
|
||||
if n % 2 == 0:
|
||||
sum = dcall(peven,n)
|
||||
else:
|
||||
sum = dcall(podd,n)
|
||||
print sum
|
|
@ -0,0 +1,21 @@
|
|||
'''
|
||||
【程序77】
|
||||
题目:填空练习(指向指针的指针)
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
main()
|
||||
{
|
||||
char *s[]={"man","woman","girl","boy","sister"};
|
||||
char **q;
|
||||
int k;
|
||||
for(k=0;k<5;k++)
|
||||
{;/*ÕâÀïÌîдʲôÓï¾ä*/
|
||||
printf("%s\n",*q);
|
||||
}
|
||||
}
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
s = ["man","woman","girl","boy","sister"]
|
||||
for i in range(len(s)):
|
||||
print s[i]
|
||||
|
|
@ -0,0 +1,14 @@
|
|||
'''
|
||||
【程序78】
|
||||
题目:找到年龄最大的人,并输出。请找出程序中有什么问题。
|
||||
1.程序分析:
|
||||
2.程序源代码
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
person = {"li":18,"wang":50,"zhang":20,"sun":22}
|
||||
m = 'li'
|
||||
for key in person.keys():
|
||||
if person[m] < person[key]:
|
||||
m = key
|
||||
|
||||
print '%s,%d' % (m,person[m])
|
|
@ -0,0 +1,18 @@
|
|||
'''
|
||||
【程序79】
|
||||
题目:字符串排序。
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
str1 = raw_input('input string:\n')
|
||||
str2 = raw_input('input string:\n')
|
||||
str3 = raw_input('input string:\n')
|
||||
print str1,str2,str3
|
||||
|
||||
if str1 > str2 : str1,str2 = str2,str1
|
||||
if str1 > str3 : str1,str3 = str3,str1
|
||||
if str2 > str3 : str2,str3 = str3,str2
|
||||
|
||||
print 'after being sorted.'
|
||||
print str1,str2,str3
|
|
@ -0,0 +1,25 @@
|
|||
'''
|
||||
【程序80】
|
||||
题目:海滩上有一堆桃子,五只猴子来分。第一只猴子把这堆桃子凭据分为五份,多了一个,这只
|
||||
猴子把多的一个扔入海中,拿走了一份。第二只猴子把剩下的桃子又平均分成五份,又多了
|
||||
一个,它同样把多的一个扔入海中,拿走了一份,第三、第四、第五只猴子都是这样做的,
|
||||
问海滩上原来最少有多少个桃子?
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
for i in range(4,10000,4):
|
||||
count = 0
|
||||
m = i
|
||||
for k in range(5):
|
||||
j = i / 4 * 5 + 1
|
||||
i = j
|
||||
if j % 4 == 0:
|
||||
count += 1
|
||||
else:
|
||||
break
|
||||
i = m
|
||||
if count == 4:
|
||||
print count
|
||||
break
|
||||
|
|
@ -0,0 +1,15 @@
|
|||
'''
|
||||
【程序81】
|
||||
题目:809*??=800*??+9*??+1 其中??代表的两位数,8*??的结果为两位数,9*??的结果为3位数。求??代表的两位数,及809*??后的结果。
|
||||
1.程序分析:
|
||||
2.程序源代码
|
||||
这个程序实在是奇怪
|
||||
0 = 1 :(
|
||||
就写个程序而已,不去追究了
|
||||
'''
|
||||
a = 809
|
||||
for i in range(10,100):
|
||||
b = i * a + 1
|
||||
if b >= 1000 and b <= 10000 and 8 * i < 100 and 9 * i >= 100:
|
||||
print b,'/',i,' = 809 * ',i,' + ', b % i
|
||||
|
|
@ -0,0 +1,12 @@
|
|||
'''
|
||||
【程序82】
|
||||
题目:八进制转换为十进制
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
n = 0
|
||||
p = raw_input('input a octal number:\n')
|
||||
for i in range(len(p)):
|
||||
n = n * 8 + ord(p[i]) - ord('0')
|
||||
print n
|
|
@ -0,0 +1,17 @@
|
|||
'''
|
||||
【程序83】
|
||||
题目:求0—7所能组成的奇数个数。
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
sum = 4
|
||||
s = 4
|
||||
for j in range(2,9):
|
||||
print sum
|
||||
if j <= 2:
|
||||
s *= 7
|
||||
else:
|
||||
s *= 8
|
||||
sum += s
|
||||
print 'sum = %d' % sum
|
|
@ -0,0 +1,31 @@
|
|||
'''
|
||||
【程序84】
|
||||
题目:一个偶数总能表示为两个素数之和。
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
此代码有问题,待修改
|
||||
'''
|
||||
import math
|
||||
if __name__ == '__main__':
|
||||
a = int(raw_input('input an odd number:\n'))
|
||||
d = 0
|
||||
c = 2
|
||||
for b in range(3,a / 2 + 1,2):
|
||||
m = 0
|
||||
for c in range(2,int(math.sqrt(b)) + 1):
|
||||
if b % c == 0 :
|
||||
m = c
|
||||
break
|
||||
|
||||
if m > math.sqrt(b):
|
||||
d = a - b
|
||||
else:
|
||||
break
|
||||
|
||||
for c in range(2,int(math.sqrt(d)) + 1):
|
||||
if d % c == 0:
|
||||
m = c
|
||||
break
|
||||
if m > math.sqrt(d):
|
||||
print '%d = %d + %d' % (a,b,d)
|
||||
|
|
@ -0,0 +1,20 @@
|
|||
'''
|
||||
【程序85】
|
||||
题目:判断一个素数能被几个9整除
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
zi = int(raw_input('input a number:\n'))
|
||||
n1 = 1
|
||||
c9 = 1
|
||||
m9 = 9
|
||||
sum = 9
|
||||
while n1 != 0:
|
||||
if sum % zi == 0:
|
||||
n1 = 0
|
||||
else:
|
||||
m9 *= 10
|
||||
sum += m9
|
||||
c9 += 1
|
||||
print '%d can be divided by %d 9' % (sum,c9)
|
|
@ -0,0 +1,13 @@
|
|||
'''
|
||||
【程序86】
|
||||
题目:两个字符串连接程序
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
a = "acegikm"
|
||||
b = "bdfhjlnpq"
|
||||
|
||||
# 连接字符串
|
||||
c = a + b
|
||||
print c
|
|
@ -0,0 +1,18 @@
|
|||
'''
|
||||
【程序87】
|
||||
题目:回答结果(结构体变量传递)
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
class student:
|
||||
x = 0
|
||||
c = 0
|
||||
def f(stu):
|
||||
stu.x = 20
|
||||
stu.c = 'c'
|
||||
a= student()
|
||||
a.x = 3
|
||||
a.c = 'a'
|
||||
f(a)
|
||||
print a.x,a.c
|
|
@ -0,0 +1,14 @@
|
|||
'''
|
||||
【程序88】
|
||||
题目:读取7个数(1—50)的整数值,每读取一个值,程序打印出该值个数的*。
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
n = 1
|
||||
while n <= 7:
|
||||
a = int(raw_input('input a number:\n'))
|
||||
while a < 1 or a > 50:
|
||||
a = int(raw_input('input a number:\n'))
|
||||
print a * '*'
|
||||
n += 1
|
|
@ -0,0 +1,23 @@
|
|||
'''
|
||||
【程序89】
|
||||
题目:某个公司采用公用电话传递数据,数据是四位的整数,在传递过程中是加密的,加密规则如下:
|
||||
每位数字都加上5,然后用和除以10的余数代替该数字,再将第一位和第四位交换,第二位和第三位交换。
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
from sys import stdout
|
||||
if __name__ == '__main__':
|
||||
a = int(raw_input('input a number:\n'))
|
||||
aa = []
|
||||
aa.append(a % 10)
|
||||
aa.append(a % 100 / 10)
|
||||
aa.append(a % 1000 / 100)
|
||||
aa.append(a / 1000)
|
||||
|
||||
for i in range(4):
|
||||
aa[i] += 5
|
||||
aa[i] %= 10
|
||||
for i in range(2):
|
||||
aa[i],aa[3 - i] = aa[3 - i],aa[i]
|
||||
for i in range(3,-1,-1):
|
||||
stdout.write(aa[i])
|
|
@ -0,0 +1,18 @@
|
|||
'''
|
||||
【程序90】
|
||||
题目:专升本一题,读结果。
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
M = 5
|
||||
a = [1,2,3,4,5]
|
||||
i = 0
|
||||
j = M - 1
|
||||
while i < M:
|
||||
a[i],a[j] = a[j],a[i]
|
||||
print a
|
||||
i += 1
|
||||
j -= 1
|
||||
for i in range(5):
|
||||
print a[i]
|
|
@ -0,0 +1,11 @@
|
|||
'''
|
||||
【程序91】
|
||||
题目:时间函数举例1
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
import time
|
||||
print time.ctime(time.time())
|
||||
print time.asctime(time.localtime(time.time()))
|
||||
print time.asctime(time.gmtime(time.time()))
|
|
@ -0,0 +1,14 @@
|
|||
'''
|
||||
【程序92】
|
||||
题目:时间函数举例2
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
import time
|
||||
start = time.time()
|
||||
for i in range(3000):
|
||||
print i
|
||||
end = time.time()
|
||||
|
||||
print end - start
|
|
@ -0,0 +1,14 @@
|
|||
'''
|
||||
【程序93】
|
||||
题目:时间函数举例3
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
import time
|
||||
start = time.clock()
|
||||
for i in range(10000):
|
||||
print i
|
||||
end = time.clock()
|
||||
print 'different is %6.3f' % (end - start)
|
||||
|
|
@ -0,0 +1,40 @@
|
|||
'''
|
||||
【程序94】
|
||||
题目:时间函数举例4,一个猜数游戏,判断一个人反应快慢。(版主初学时编的)
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
import time
|
||||
import random
|
||||
|
||||
play_it = raw_input('do you want to play it.(\'y\' or \'n\')')
|
||||
while play_it == 'y':
|
||||
c = raw_input('input a character:\n')
|
||||
i = random.randint(0,2**32) % 100
|
||||
print 'please input number you guess:\n'
|
||||
start = time.clock()
|
||||
a = time.time()
|
||||
guess = int(raw_input('input your guess:\n'))
|
||||
while guess != i:
|
||||
if guess > i:
|
||||
print 'please input a little smaller'
|
||||
guess = int(raw_input('input your guess:\n'))
|
||||
else:
|
||||
print 'please input a little bigger'
|
||||
guess = int(raw_input('input your guess:\n'))
|
||||
end = time.clock()
|
||||
b = time.time()
|
||||
var = (end - start) / 18.2
|
||||
print var
|
||||
# print 'It took you %6.3 seconds' % time.difftime(b,a))
|
||||
if var < 15:
|
||||
print 'you are very clever!'
|
||||
elif var < 25:
|
||||
print 'you are normal!'
|
||||
else:
|
||||
print 'you are stupid!'
|
||||
print 'Congradulations'
|
||||
print 'The number you guess is %d' % i
|
||||
play_it = raw_input('do you want to play it.')
|
||||
|
|
@ -0,0 +1,11 @@
|
|||
'''
|
||||
【程序96】
|
||||
题目:计算字符串中子串出现的次数
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
str1 = raw_input('input a string:\n')
|
||||
str2 = raw_input('input a sub string:\n')
|
||||
ncount = str1.count(str2)
|
||||
print ncount
|
|
@ -0,0 +1,17 @@
|
|||
'''
|
||||
【程序97】
|
||||
题目:从键盘输入一些字符,逐个把它们送到磁盘上去,直到输入一个#为止。
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
from sys import stdout
|
||||
filename = raw_input('input a file name:\n')
|
||||
fp = open(filename,"w")
|
||||
ch = raw_input('input string:\n')
|
||||
while ch != '#':
|
||||
fp.write(ch)
|
||||
stdout.write(ch)
|
||||
ch = raw_input('')
|
||||
fp.close()
|
||||
|
|
@ -0,0 +1,15 @@
|
|||
'''
|
||||
【程序98】
|
||||
题目:从键盘输入一个字符串,将小写字母全部转换成大写字母,然后输出到一个磁盘文件“test”中保存。
|
||||
输入的字符串以!结束。
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
fp = open('test.txt','w')
|
||||
string = raw_input('please input a string:\n')
|
||||
string = string.upper()
|
||||
fp.write(string)
|
||||
fp = open('test.txt','r')
|
||||
print fp.read()
|
||||
fp.close()
|
|
@ -0,0 +1,25 @@
|
|||
'''
|
||||
程序99】
|
||||
题目:有两个磁盘文件A和B,各存放一行字母,要求把这两个文件中的信息合并(按字母顺序排列),
|
||||
输出到一个新文件C中.
|
||||
1.程序分析:
|
||||
2.程序源代码:
|
||||
'''
|
||||
if __name__ == '__main__':
|
||||
import string
|
||||
fp = open('JCP099.py')
|
||||
a = fp.read()
|
||||
fp.close()
|
||||
|
||||
fp = open('JCP098.py')
|
||||
b = fp.read()
|
||||
fp.close()
|
||||
|
||||
fp = open('C.txt','w')
|
||||
l = list(a + b)
|
||||
l.sort()
|
||||
s = ''
|
||||
s = s.join(l)
|
||||
fp.write(s)
|
||||
fp.close()
|
||||
|
Loading…
Reference in New Issue