HDU-6441-Find Integer（费马大定理+勾股数）ʕ •ᴥ•ʔ

Problem Description

people in USSS love math very much, and there is a famous math problem .

give you two integers n ,a ,you are required to find 2 integers b ,c such that an +bn=cn .

Input

one line contains one integer T ;(1≤T≤1000000)

next T lines contains two integers n ,a ;(0≤n≤1000 ,000 ,000,3≤a≤40000)

Output

print two integers b ,c if b ,c exits;(1≤b,c≤1000 ,000 ,000) ;

else print two integers -1 -1 instead.

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Sample Input

1 2 3

Sample Output

4 5

题目大意：给出T组测试样例，对于每组测试数据，输入n，a，输出满足a^n+b^n=c^n的b和c，如果没有满足要求的，输出-1 -1；

费马大定理证明了n>2的时候没有整数满足a^n+b^n==c^n；

当n==1时，任意a，1，a+1即可；

当n==2时，即为勾股数，百度百科有证明过程，分为奇数和偶数，然后根据式子直接输出结果即可：

当a为大于1的奇数2n+1时，b=2n^2+2n, c=2n^2+2n+1

当a为大于4的偶数2n时，b=n^2-1, c=n^2+1

#include <cstring>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <string.h>
//#include <map>
#define ll long long
using namespace std;
#define pai acos(-1,0)
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		ll n,a;
		ll b,c;
		scanf("%lld %lld",&n,&a);
		if(n>2)
		printf("-1 -1\n");
		if(n==1) 
		printf("1 %lld\n",a+1);
		if(n==2)
		{
			if(a%2)
			{
				b=(a-1)*(a-1)/2+a-1;
				c=b+1;
			}
			else
			{
				b=a*a/4-1;
				c=b+2;
			}
			printf("%lld %lld\n",b,c); 
		}
		
	}
	return 0;
}