2018中国大学生程序设计竞赛

Find Integer

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0
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 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

由费马大定理知n>=3时必然无解，因此只需讨论n==2的情况

参见：http://tieba.baidu.com/p/108936816

code:

#include<bits/stdc++.h>
#include<stdio.h>
#include<iostream>
#include<cmath>
#include<math.h>
#include<queue>
#include<set>
#include<map>
#include<iomanip>
#include<algorithm>
#include<stack>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
ll b[40005];
ll c[40005];
int id=1;
ll num[10005];
void init()
{
    for(ll aa=3;aa<=40000;aa++)
    {
        if(aa%2)
        {
            b[aa]=(aa/2)*(aa/2)+(aa/2+1)*(aa/2+1)-1;
            c[aa]=(aa/2)*(aa/2)+(aa/2+1)*(aa/2+1);
        }
        else
        {
            b[aa]=(aa/2)*(aa/2)-1;
            c[aa]=(aa/2)*(aa/2)+1;
        }
    }
}
int main()
{
#ifndef ONLINE_JUDGE
    //freopen("in.txt","r",stdin);
#endif // ONLIN
init();

    int t;
    cin>>t;
    ll a,n;
    while(t--)
    {
        scanf("%lld%lld",&n,&a);
        if(n>=3)
        {
            printf("-1 -1
");continue;
        }
        else if(n==0)
        {
           printf("-1 -1
");
        }
        else if(n==1)
        {
            printf("1 %lld
",a+1);
        }
        else if(n==2)
        {
            printf("%lld %lld
",b[a],c[a]);
        }
    }

return 0;
}