(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

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

 1 10
 2 4 4
 3 5 6
 4 10 10
 5 10 25
 6 20000 100000000
 7 0 5
 8 3 6
 9 220 284
10 0 1
11 1000000000 1000000000

输出

 1 2 2
 2 2 3
 3 -1 -1
 4 5 5
 5 10000 10000
 6 474848249 525151758
 7 352077071 647922939
 8 448762649 551237578
 9 -1 -1
10 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：

 1 #include<bits/stdc++.h>
 2 typedef long long ll;
 3 using namespace std;
 4 const int mod=1e9+7;
 5 ll pow_mod(ll a, ll b){
 6     ll ret = 1;
 7     while(b){
 8         if(b & 1) ret = (ret * a) % mod;
 9         a = (a * a) % mod;
10         b >>= 1;
11     }
12     return ret;
13 }
14 int main() {
15     ll t,b,c,x,y,a;
16     scanf("%lld",&t);
17     while(t--)
18     {
19         scanf("%lld%lld",&b,&c);
20         int now=(b*b-4*c+mod)%mod;
21         if(pow_mod(now,(mod-1)/2)==mod-1)
22         {
23             printf("-1 -1
");continue;
24         }
25         a=pow_mod(now,(mod+1)/4);
26         x=(((a+b)*(mod+1)/2)%mod+mod)%mod;
27         y=(((b-a)*(mod+1)/2)%mod+mod)%mod;
28         if(x>y) printf("%lld %lld
",y,x);
29         else  printf("%lld %lld
",x,y);
30     }
31     return 0;
32 }