URAL 1349. Farm （费马大定理）

1349. Farm

Time limit: 1.0 second

Memory limit: 64 MB

Here is a farm. Here is a farmer that counts how many animal live in his farm: a camels, b sheep, c green cockroaches. Occurs that an +
bn = cn. n is given. You are to find all the rest.

Input

n (0 ≤ n ≤ 100)

Output

Three different integers (a, b and c) such that an + bn =
cn, 1 ≤ a, b, c ≤ 100. If there are several solutions you should output the one where a is minimal. If there are several solutions with the minimal a you should output the one with
minimal b, and so on. Output −1 if there is no solution.

Samples

input	output
0	-1
1	1 2 3

题意：给定n找到a，b和c满足，a^n + b^n = c^n.

解析：根据费马大定理，当n>=时，无解，故直接枚举即可。

AC代码：

#include <cstdio>

int main(){
    int n;
    while(scanf("%d", &n)==1){
        if(n == 1) puts("1 2 3");
        else if(n == 2) puts("3 4 5");
        else puts("-1");
    }
    return 0;
}

费马大定理，真好^_^