(x+y)%p==b ;(x*y)%p==c; 知道b,c，求x,y;

 题目如下：

来源：牛客网

时间限制：C/C++ 2秒，其他语言4秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld

题目描述

Amy asks Mr. B  problem B. Please help Mr. B to solve the following problem.

Let p = 1000000007.

Given two integers b and c, please find two integers x and  y (0≤x≤y<p) such that

(x+y) mod p==b    

(x*y) mod p==c

输入描述:

The first line contains an integer t, which is the number of test cases (1 <= t <= 10).

In the following t lines, each line contains two integers b and c (0 <= b, c < p).

输出描述:

For each test case, please output x, y in one line.
If there is a solution, because x <= y, the solution is unique.

If there is no solution, please output -1, -1

示例1

输入

10
4 4
5 6
10 10
10 25
20000 100000000
0 5
3 6
220 284
0 1
1000000000 1000000000

输出

2 2
2 3
-1 -1
5 5
10000 10000
474848249 525151758
352077071 647922939
448762649 551237578
-1 -1
366417496 633582504

---

题解：公式法

• 这个题目是某次翻Wikipedia的时候想到的
• https://en.wikipedia.org/wiki/Quadratic_residue
• 解二次方程，核心在于求二次剩余（求平方根）
• 如果p % 4 =3，x^2 % p = a
• 那么x = ±pow(a, (p+1)/4, p)
• 然后就和一般的方程一样解就可以了
• 判断是否有解用
• https://en.wikipedia.org/wiki/Euler%27s_criterion

上代码：

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
const int mod=1e9+7;
ll pow_mod(ll a, ll b){
    ll ret = 1;
    while(b){
        if(b & 1) ret = (ret * a) % mod;
        a = (a * a) % mod;
        b >>= 1;
    }
    return ret;
}
int main() {
    ll t,b,c,x,y,a;
    scanf("%lld",&t);
    while(t--)
    {
        scanf("%lld%lld",&b,&c);
        int now=(b*b-4*c+mod)%mod;
        if(pow_mod(now,(mod-1)/2)==mod-1)
        {
            printf("-1 -1
");continue;
        }
        a=pow_mod(now,(mod+1)/4);
        x=(((a+b)*(mod+1)/2)%mod+mod)%mod;
        y=(((b-a)*(mod+1)/2)%mod+mod)%mod;
        if(x>y) printf("%lld %lld
",y,x);
        else  printf("%lld %lld
",x,y);
    }
    return 0;
}