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<>问题
公鸡五块钱一只，母鸡三块钱一只，小鸡一块钱三只，现在要用一百块钱买一百只鸡，问公鸡、母鸡、小鸡各多少只？总共有多少种方案？
<>思路
<>有题可得两个条件：
1.公鸡X5 + 母鸡X3 + 小鸡/3 = 100
2.公鸡 + 母鸡 + 小鸡 =100
所以编写程序的时候需要满足这个两个条件
<>分析方案
只买公鸡，方案最多有二十种，只买母鸡方案最多有三十三种，只买小鸡方案最多有一百种，但是编写程序的时候因为含头不含尾，所以写程序的时候要多加一
<>编写程序
<>程序一、采用三重循环解决
count = 0 for cock in range(21): for hen in range(35): for chick in range(101):
if cock + hen + chick == 100 and cock * 5 + hen * 3 + chick / 3 == 100: count =
count+ 1 print('公鸡：{}\t 母鸡:{}\t 小鸡:{}\t'.format(cock,hen,chick)) print(
'统计结果：百钱买百鸡总共有{}种方案。'.format(count))
<>运行结果
<>程序二、采用双重循环解决
由于知道购买鸡的总数量，知道了公鸡和母鸡的数量，小鸡的数量也就自然而然的确定了，所以请看下面程序
count = 0 for cock in range(21): for hen in range(35): chick = 100 - cock - hen
if cock + hen + chick == 100： count = count + 1 print('公鸡：{}\t 母鸡:{}\t 小鸡:{}\t'.
format(cock,hen,chick)) print('统计结果：百钱买百鸡总共有{}种方案。'.format(count))
<>运行结果
结果还是一样的，而且第二种方案更简单更简便
