费马大定理

Posted 2021-02-17 ylrwj

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 .Inputone 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)
Outputprint 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
费马大定理，当n>2 时 a^n+b^n=c^n 无解
所以可以分为三种情况当n=0 题目说了 b和c必须>=1 
n=1 b。c 解不唯一
当n=2 就是勾股定理了---->勾股数组
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<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int main()
{
    int t;
    long long n,a,b,c;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld%lld",&n,&a);
        if(n==0)
           printf("-1 -1
");
        else if(n==1)
           printf("%lld %lld",a+1,2*a+1);
        else if(n==2)
        {
            if(a%2==1)
            {
                long long x=(a-1)/2;
                b=2*x*x+2*x;
                c=2*x*x+2*x+1;
                printf("%lld %lld
",b,c);
            }
            if(a%2==0)
            {
                int x=a/2;
                b=x*x-1;
                c=x*x+1;
                printf("%lld %lld
",b,c);
            }
        }
        else
          printf("-1 -1
");
    }
    return 0;
}