有一群鸡和一群兔，它们的只数相同，它们的脚数都是三位数，且这两个三位数的数字分别是0，1，2，3，4，5。设计一个算法用蛮力法求鸡和兔各有多少只，?它们的脚数各是多少

102/2=51可知只数>=51
542/4=135可知只数<=135
循环只数从51到135 计算鸡脚数A和兔脚数B   
判断A*1000+B是否有重复数字即可

from itertools import permutations
feet = permutations([0, 1, 2, 3, 4, 5], 3)
feet = list(set([int(''.join([str(j) for j in i])) for i in feet if len(i) == 3 and i[0] != 0]))
for chicken in feet:
    for rabbit in feet:
        if chicken*2 == rabbit and (chicken%2 == 0):
            if len(set(str(chicken))^set(str(rabbit))) == 6:
                print(f"雞 {chicken>>1} 只，{chicken} 只腳\n兔 {rabbit>>2} 只，{rabbit } 只腳")

輸出：

雞 76 只，152 只腳
兔 76 只，304 只腳

[此贴子已经被作者于2021-8-29 00:25编辑过]

void main()
{
    int i, j, k;
    int v1, v2, tmp;

    for (i=1; i<3; i++)
    {
        for (j=0; j<6; j++)
        {
            for (k=0; k<6; k++)
            {
                v1 = i*100 + j*10 + k;
                v2 = v1*2;
                tmp = (1<<i) | (1<<j) | (1<<k);
                tmp |= (1<<(v2/100)) | (1<<((v2/10)%10)) | (1<<(v2%10));

                if ((v1 & 1)==0 && tmp==0x3F)
                {
                    printf("%d %d\n", v1, v2);
                }
            }
        }
    }
}

[此贴子已经被作者于2021-9-9 16:55编辑过]

for chick in range(102,255,2):
    rabbit=2*chick
    if all((str(chick)+str(rabbit)).count(d) for  d in "012345"):
        print(f"鸡 {chick//2} 只，{chick} 只脚;兔 {rabbit//4} 只，{rabbit} 只脚")

[此贴子已经被作者于2022-12-17 22:13编辑过]