论坛 › Python交流 › 韩信点兵问题

初学者8 发表于 2022-10-25 09:34:14

韩信点兵问题

韩信有一队兵，他想知道有多少人，便让士兵排队报数：按从1至5报数，最末一个士兵报的数为1；按从1至6报数，最末一个士兵报的数为5；按从1至7报数，最末一个士兵报的数为4；最后再按从1至11报数，最末一个士兵报的数为10。编程求韩信至少有多少兵？用for循环怎么搞{:10_266:}

jackz007 发表于 2022-10-25 10:09:07

n = 1
while True:
    if n % 5 == 1 and n % 6 == 5 and n % 7 == 4 and n % 11 == 10:
      print(n)
      break
    n += 1
      运行实况：
D:\\Python>python x.py
2111

D:\\Python>

人造人 发表于 2022-10-25 10:14:55

答案是多少？

sh-5.1$ cat main.py
#!/usr/bin/env python
#coding=utf-8

count = 0
while True:
    a = (count % 5) + 1
    b = (count % 6) + 1
    c = (count % 7) + 1
    d = (count % 11) + 1
    count += 1
    if a != 1: continue
    if b != 5: continue
    if c != 4: continue
    if d != 10: continue
    print(count)
    break
sh-5.1$ ./main.py
2111
sh-5.1$


sh-5.1$ cat main.py
#!/usr/bin/env python
#coding=utf-8

from itertools import repeat

count = 0
for _ in repeat(None):
    a = (count % 5) + 1
    b = (count % 6) + 1
    c = (count % 7) + 1
    d = (count % 11) + 1
    count += 1
    if a != 1: continue
    if b != 5: continue
    if c != 4: continue
    if d != 10: continue
    print(count)
    break
sh-5.1$ ./main.py
2111
sh-5.1$


sh-5.1$ cat main.py
#!/usr/bin/env python
#coding=utf-8

from itertools import count

for i in count():
    a = (i % 5) + 1
    b = (i % 6) + 1
    c = (i % 7) + 1
    d = (i % 11) + 1
    if a != 1: continue
    if b != 5: continue
    if c != 4: continue
    if d != 10: continue
    print(i)
    break
sh-5.1$ ./main.py
2110
sh-5.1$

人造人 发表于 2022-10-25 10:15:29

2110是吗？

人造人 发表于 2022-10-25 10:18:13

sh-5.1$ cat main.py
#!/usr/bin/env python
#coding=utf-8

count = -1
while True:
    count += 1
    a = (count % 5) + 1
    b = (count % 6) + 1
    c = (count % 7) + 1
    d = (count % 11) + 1
    if a != 1: continue
    if b != 5: continue
    if c != 4: continue
    if d != 10: continue
    print(count)
    break
sh-5.1$ ./main.py
2110
sh-5.1$


sh-5.1$ cat main.py
#!/usr/bin/env python
#coding=utf-8

from itertools import repeat

count = -1
for _ in repeat(None):
    count += 1
    a = (count % 5) + 1
    b = (count % 6) + 1
    c = (count % 7) + 1
    d = (count % 11) + 1
    if a != 1: continue
    if b != 5: continue
    if c != 4: continue
    if d != 10: continue
    print(count)
    break
sh-5.1$ ./main.py
2110
sh-5.1$

人造人 发表于 2022-10-25 10:20:15

这个版本的最好，我最喜欢
sh-5.1$ cat main.py
#!/usr/bin/env python
#coding=utf-8

from itertools import count

for i in count():
    a = (i % 5) + 1
    b = (i % 6) + 1
    c = (i % 7) + 1
    d = (i % 11) + 1
    if a != 1: continue
    if b != 5: continue
    if c != 4: continue
    if d != 10: continue
    print(i)
    break
sh-5.1$ ./main.py
2110
sh-5.1$

人造人 发表于 2022-10-25 10:24:41

答案是 2111
sh-5.1$ cat main.py
#!/usr/bin/env python
#coding=utf-8

from itertools import count

for i in count():
    a = (i % 5) + 1
    b = (i % 6) + 1
    c = (i % 7) + 1
    d = (i % 11) + 1
    if a != 1: continue
    if b != 5: continue
    if c != 4: continue
    if d != 10: continue
    print(i + 1)
    break
sh-5.1$ ./main.py
2111
sh-5.1$

查看完整版本: 韩信点兵问题