HDU 6441 Find Integer（费马大定理）

Find Integer

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1199    Accepted Submission(s): 340
Special Judge
 

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.

Sample Input

1 2 3

Sample Output

4 5

Source

​​2018中国大学生程序设计竞赛 - 网络选拔赛​​

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费马大定理了解一下，（我好像听老师提起过）

这个题 用 cin cout 就超时 ，无语 - -

/*
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 所有 无解
当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 <bits/stdc++.h>
#define ll long long
using namespace std;

int main()
{
    //ios::sync_with_stdio(false); 用 cin cout 就 TLE
    int t;
    ll n,a;
    scanf("%d",&t);
    while(t--)
    {
        ll b,c;
        scanf("%lld%lld",&n,&a);
        if(n==0) printf("-1 -1\n");
        else if(n==1) printf("%lld %lld\n",a+1,2*a+1);
        else if(n==2) {
            if(a%2==1) //奇数
            {
               ll nn=(a-1)/2;
               b=2*nn*nn+2*nn;
               c=2*nn*nn+2*nn+1;
               printf("%lld %lld\n",b,c);
            }
            else if(a%2==0)
            {
                ll nn=a/2;
                b=nn*nn-1;
                c=nn*nn+1;
                printf("%lld %lld\n",b,c);
            }
        }else printf("-1 -1\n");
    }return 0;
}