HDUOJ 6441 Find Integer

HDUOJ 6441 Find Integer

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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 $a^n+b^n=c^n$.

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.

Sample Input

1
2 3

Sample Output

4 5

Thinking questions~
First, according to Fermat's Last Theorem, when $n>When\; 2$ , there must be no positive integer solution ~
when$n=At\; 0$ , there is obviously no solution ~
when$n=When\; 1$ , output$1$ sum$a+1\; is$ enough,
when$n=At\; 2$ o'clock, note that the title limit must be a positive integer, we consider parity, for$a$ is odd,$a^2$ can be split into$a^2\ast 1$ ; to$a$ is an even number,$a^2$ can be split into$\frac{a^2}{2}\ast 2.$ Solve a binary linear equation separately~
pay attention to the output order, the AC code is as follows:

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int T,n,a;
int main(){
    
    
    scanf("%d",&T);
    while(T--){
    
    
        scanf("%d%d",&n,&a);
        if(n>2||n==0) printf("-1 -1\n");
        else if(n==2){
    
    
            if(a%2) printf("%d %d\n",(a*a-1)/2,(a*a+1)/2);
            else printf("%d %d\n",a*a/4-1,a*a/4+1);
        }else if(n==1){
    
    
            printf("1 %d\n",a+1);
        }
    }
    return 0;
}