URAL 1349 Farm

Discription

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 a ⁿ + b ⁿ = c ⁿ. 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 a ⁿ + b ⁿ = c ⁿ, 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.

Example

input	output
0	-1
1	1 2 3

以前一直吐槽费马大定理实在是废，结果今天终于碰着了hhhhh。

这个定理贼简单，大致就是对于n>=3,不存在整数a,b,c使得a^n+b^n=c^n。

然后这就是个水题了hhhhh

(我直接在提交界面写的代码都没编译hhhhh)

#include<cstdio>

#include<cstdlib>

#include<algorithm>

#include<cmath>

using namespace std;

int n;

int main(){

    scanf("%d",&n);

    if(!n||n>) puts("-1");

    else{

        if(n==) puts("1 2 3");

        else puts("3 4 5");

    }

    return ;

}

巴特西

URAL 1349 Farm

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