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python 找零钱
按照题目要求编写的Python找零钱程序如下
def smallchange(money):
fifty=50
ten=10
five=5
one=1
print(money//fifty,end=' ')
money=money%fifty
print(money//ten,end=' ')
money=money%ten
print(money//five,end=' ')
money=money%five
print(money//one)
l=[]
n=int(input())
for i in range(n):
l.append(input())
for i in range(n):
pay,payable=l[i].split(' ')
money=int(pay)-int(payable)
smallchange(money)
源代码(注意源代码的缩进)
解释一下python 的一个找零钱程序要很详细。求大神
cate是一个元组，用来记录人名币面额 100 0.1 分别代表面额，在这个程序里面必须得按照从大到小的面额显示,不然显示就会有问题。
sy是个变量，用来记录钱数的。
jine 是一个变量，用来记录当前循环面额大小的(这里乘以了10，是便于计算才这么做的)
tmpStr用来记录已经统计的钱数的。59153.8/100=591 统计100元有591张
tmp是个列表，用来保存最后所有统计数据的
tmpUn 是用来确定计算时候是使用单位元还是单位角的
int 这个是用来将字符串变为int类型的操作 本来是字符串'123' 变为数字123
用python 怎么 解 将100元兑换为1元、5、10元的零钱，请问有多少种兑换方法？
cnt = 0
for i in xrange(100 / 1 + 1):
for j in xrange((100 - i * 1) / 5 + 1):
for k in xrange ((100 - i * 1 + j * 5) / 10 + 1):
if i * 1 + j * 5 + k * 10 == 100:
print "1:", i, "5:", j, "10:", k
cnt += 1
print cnt
结果如下，总共121种
1: 0 5: 0 10: 10
1: 0 5: 2 10: 9
1: 0 5: 4 10: 8
1: 0 5: 6 10: 7
1: 0 5: 8 10: 6
1: 0 5: 10 10: 5
1: 0 5: 12 10: 4
1: 0 5: 14 10: 3
1: 0 5: 16 10: 2
1: 0 5: 18 10: 1
1: 0 5: 20 10: 0
1: 5 5: 1 10: 9
1: 5 5: 3 10: 8
1: 5 5: 5 10: 7
1: 5 5: 7 10: 6
1: 5 5: 9 10: 5
1: 5 5: 11 10: 4
1: 5 5: 13 10: 3
1: 5 5: 15 10: 2
1: 5 5: 17 10: 1
1: 5 5: 19 10: 0
1: 10 5: 0 10: 9
1: 10 5: 2 10: 8
1: 10 5: 4 10: 7
1: 10 5: 6 10: 6
1: 10 5: 8 10: 5
1: 10 5: 10 10: 4
1: 10 5: 12 10: 3
1: 10 5: 14 10: 2
1: 10 5: 16 10: 1
1: 10 5: 18 10: 0
1: 15 5: 1 10: 8
1: 15 5: 3 10: 7
1: 15 5: 5 10: 6
1: 15 5: 7 10: 5
1: 15 5: 9 10: 4
1: 15 5: 11 10: 3
1: 15 5: 13 10: 2
1: 15 5: 15 10: 1
1: 15 5: 17 10: 0
1: 20 5: 0 10: 8
1: 20 5: 2 10: 7
1: 20 5: 4 10: 6
1: 20 5: 6 10: 5
1: 20 5: 8 10: 4
1: 20 5: 10 10: 3
1: 20 5: 12 10: 2
1: 20 5: 14 10: 1
1: 20 5: 16 10: 0
1: 25 5: 1 10: 7
1: 25 5: 3 10: 6
1: 25 5: 5 10: 5
1: 25 5: 7 10: 4
1: 25 5: 9 10: 3
1: 25 5: 11 10: 2
1: 25 5: 13 10: 1
1: 25 5: 15 10: 0
1: 30 5: 0 10: 7
1: 30 5: 2 10: 6
1: 30 5: 4 10: 5
1: 30 5: 6 10: 4
1: 30 5: 8 10: 3
1: 30 5: 10 10: 2
1: 30 5: 12 10: 1
1: 30 5: 14 10: 0
1: 35 5: 1 10: 6
1: 35 5: 3 10: 5
1: 35 5: 5 10: 4
1: 35 5: 7 10: 3
1: 35 5: 9 10: 2
1: 35 5: 11 10: 1
1: 35 5: 13 10: 0
1: 40 5: 0 10: 6
1: 40 5: 2 10: 5
1: 40 5: 4 10: 4
1: 40 5: 6 10: 3
1: 40 5: 8 10: 2
1: 40 5: 10 10: 1
1: 40 5: 12 10: 0
1: 45 5: 1 10: 5
1: 45 5: 3 10: 4
1: 45 5: 5 10: 3
1: 45 5: 7 10: 2
1: 45 5: 9 10: 1
1: 45 5: 11 10: 0
1: 50 5: 0 10: 5
1: 50 5: 2 10: 4
1: 50 5: 4 10: 3
1: 50 5: 6 10: 2
1: 50 5: 8 10: 1
1: 50 5: 10 10: 0
1: 55 5: 1 10: 4
1: 55 5: 3 10: 3
1: 55 5: 5 10: 2
1: 55 5: 7 10: 1
1: 55 5: 9 10: 0
1: 60 5: 0 10: 4
1: 60 5: 2 10: 3
1: 60 5: 4 10: 2
1: 60 5: 6 10: 1
1: 60 5: 8 10: 0
1: 65 5: 1 10: 3
1: 65 5: 3 10: 2
1: 65 5: 5 10: 1
1: 65 5: 7 10: 0
1: 70 5: 0 10: 3
1: 70 5: 2 10: 2
1: 70 5: 4 10: 1
1: 70 5: 6 10: 0
1: 75 5: 1 10: 2
1: 75 5: 3 10: 1
1: 75 5: 5 10: 0
1: 80 5: 0 10: 2
1: 80 5: 2 10: 1
1: 80 5: 4 10: 0
1: 85 5: 1 10: 1
1: 85 5: 3 10: 0
1: 90 5: 0 10: 1
1: 90 5: 2 10: 0
1: 95 5: 1 10: 0
1: 100 5: 0 10: 0
121
求解答python，编写程序1元人民币换成1分、2分、5分有多少种方案，并展示出来是哪些方案
刚刚回答错了，分看成了角，下面是正确的
money = 1
count = 0
print('\t\t1分\t2分\t5分\(张)')
def jihua(money):
global count
for i in range(100):
for j in range(100):
for n in range(100):
if i * 0.01 + j * 0.02 + n * 0.05 == money:
count += 1
print('方案%d\t%d\t%d\t%d'
% (count, i, j, n))
jihua(money)
print('有%d种方案。' % count)
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