python之捕鱼问题

2018.03.04
auther:wills

捕鱼，有abcde五个人去捕鱼，补了很多鱼fish，休息一夜，a最先醒来，扔了一条鱼然后平均分成五份，拿走了自己的一份。b第二个醒来，也扔了一条鱼然后平均分成五份，拿走了自己的一份。c第三个醒来，d第四个醒来，e第五个醒来均进行了相同的动作，问：最少有多少条鱼？

先从正向捕鱼说起

下面是我最初写的代码

fish = 1
while True
    flag = 1
    for _ in range(5):
        if (fish - 1)%5==0:
            fish = (fish - 1)//5*4
        else:
            flag = 0
            break
    if flag:
        print(fish)
        break
    fish += 1

这段代码我已运行就进入死循环，找了半天都没找到问题点，最后才发现我从fish=1开始，进入if语句后fish被修改为0，然后又fish+=1，进入了死循环。修改后代码如下

fish = 1
while 1:
    flag = 1
    total = fish
    for _ in range(5):
        if (total - 1)%5==0:
            total = (total -1)//5*4
        else:
            flag = 0
    if flag:
        print(fish)
        break
    fish += 1

逆向思维，假设第五个人分鱼后剩下的鱼为fish，这样分鱼过程就只有4次，可以少一个循环代码如下

for fish in range(1,10000):
    for n in range(4)：
        if (fish -1 )%5 == 0 and fish%4 == 0：
            fish = fish//4*5+1
            if n ==3:
                print(fish)
       # else:
       #     break

这个分鱼方法有个好处，就是代码极少，只有6行，#号注释的两行只是用于跳出for循环提高代码执行速度

下面这个while＋for循环的办法可以更高的提高代码执行速度，不过稍微需要多加几个变量

fish = 1
flag = 1
while flag:
    total = fish
    for n in range(4):
        if (total-1)%5==0 and total%4==0:
            total = total//4*5+1
            if n==3:
                print(fish)
                print(total)
                flag =0
    fish += 1

下面我把4种捕鱼代码都放到一起，并且使用了一个time函数来进行计时，比较几种代码的运行效率

from time import time
a1 = time()
fish = 1
while 1:
    flag = 1
    total = fish
    for _ in range(5):
        if (total - 1)%5==0:
            total = (total -1)//5*4
        else:
            flag = 0
    if flag:
        print(fish)
        break
    fish += 1
a2 = time()
for fish in range(1,10000):
    for n in range(4):
        if (fish -1 )%5 == 0 and fish%4 == 0:
            fish = fish//4*5+1
            if n ==3:
                print(fish)
a3 = time()
for fish in range(1,10000):
    for n in range(4):
        if (fish -1 )%5 == 0 and fish%4 == 0:
            fish = fish//4*5+1
            if n ==3:
                print(fish)
        else:
            break
a4 = time()

fish = 1
flag = 1
while flag:
    total = fish
    for n in range(4):
        if (total-1)%5==0 and total%4==0:
            total = total//4*5+1
            if n==3:
                print(fish)
                print(total)
                flag =0
    fish += 1

a5 = time()
print(a2-a1)
print(a3-a2)
print(a4-a3)
print(a5-a4)