nefu702 The minimum square sum(数论)

The minimum square sum

Time Limit 1000ms Memory Limit 65536K

description

Given a prime p (p<108),you are to find min{x2+y2},where x and y belongs to positive integer, so that x2+y2=0 (mod p).

input

Every line is a p. No more than 10001 test cases.

output

The minimum square sum as described above.

sample_input

2 3 5

sample_output

2 18 5

hint

source

思路：结论p=4k+3则没有平方和会等于P，否则就有。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>

using namespace std;

int main()
{
    long long p;
    while (cin>>p)
    {
        if (p==2) cout << "2" << endl;
        else if((p-3)%4==0) cout << 2*p*p << endl;
        else cout << p << endl;
    }
    return 0;
}