题干：

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

give you two integers nn,aa,you are required to find 22 integers bb,cc such that anan+bn=cnbn=cn.

Input

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

next TT lines contains two integers nn,aa;(0≤n≤1000(0≤n≤1000,000000,000,3≤a≤40000)000,3≤a≤40000) 

Output

print two integers bb,cc if bb,cc exits;(1≤b,c≤1000(1≤b,c≤1000,000000,000)000); 

else print two integers -1 -1 instead.

Sample Input


1 2 3

Sample Output


4 5

解题报告：

        根据费马大定理的结论，用奇偶构造法求出勾股定理中另外两个数。

AC代码：

#include<bits/stdc++.h>

using namespace std;

int main()
{
  int t;
  cin>>t;
  while(t--) {
    int n,a;
    scanf("%d%d",&n,&a);
    if(n > 2 || n==0) {
      puts("-1 -1");
    }
    else if(n == 1) {
      printf("1 %d\n",a+1);
    }
    else {
      if(a&1) {
        int tmp = (a-1)/2;
        printf("%d %d\n",2*tmp*tmp + 2*tmp,2*tmp*tmp+2*tmp+1);
      }
      else {
        int tmp = a/2 - 1;
        printf("%d %d\n",tmp*tmp + 2*tmp,tmp*tmp + 2*tmp + 2);
      }
    }
  }
  
  return 0 ;
}

 

知识点：（勾股定理的构造）

【HDU - 6441】Find Integer （费马大定理 + 奇偶数列法构造勾股定理）_#include