URAL 1132 Square Root（二次剩余定理）题解

题意：

求\(x^2 \equiv a \mod p\) 的所有整数解

思路：

二次剩余定理求解。

参考：

二次剩余Cipolla's algorithm学习笔记

板子：

//二次剩余，p是奇质数
ll ppow(ll a, ll b, ll mod){
    ll ret = 1;
    a = a % mod;
    while(b){
        if(b & 1) ret = ret * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ret;
}
struct TT{
    ll p, d;
};
ll w;
TT mul_er(TT a, TT b, ll mod){
    TT ans;
    ans.p = (a.p * b.p % mod + a.d * b.d % mod * w % mod) % mod;
    ans.d = (a.p * b.d % mod + a.d * b.p % mod) % mod;
    return ans;
}
TT power(TT a, ll b, ll mod){
    TT ret;
    ret.p = 1, ret.d = 0;
    while(b){
        if(b & 1) ret = mul_er(ret, a, mod);
        a = mul_er(a, a, mod);
        b >>= 1;
    }
    return ret;
}
ll legendre(ll a, ll p){
    return ppow(a, (p - 1) >> 1, p);
}
ll modulo(ll a, ll mod){
    a %= mod;
    if(a < 0) a += mod;
    return a;
}
ll solve(ll n, ll p){   //x^2 = n mod p
    if(n == 0) return 0;
    if(n == 1) return 1;
    if(p == 2) return 1;
    if(legendre(n, p) + 1 == p) return -1;  //无解
    ll a = -1, t;
    while(true){
        a = rand() % p;
        t = a * a - n;
        w = modulo(t, p);
        if(legendre(w, p) + 1 == p) break;
    }
    TT temp;
    temp.p = a;
    temp.d = 1;
    TT ans = power(temp, (p + 1) >> 1, p);
    return ans.p;
}

代码：

#include<map>
#include<set>
#include<queue>
#include<stack>
#include<ctime>
#include<cmath>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 5e4 + 5;
const int INF = 0x3f3f3f3f;
const ull seed = 131;
const ll MOD = 1e9 + 7;
using namespace std;

//二次剩余
ll ppow(ll a, ll b, ll mod){
    ll ret = 1;
    a = a % mod;
    while(b){
        if(b & 1) ret = ret * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return ret;
}
struct TT{
    ll p, d;
};
ll w;
TT mul_er(TT a, TT b, ll mod){
    TT ans;
    ans.p = (a.p * b.p % mod + a.d * b.d % mod * w % mod) % mod;
    ans.d = (a.p * b.d % mod + a.d * b.p % mod) % mod;
    return ans;
}
TT power(TT a, ll b, ll mod){
    TT ret;
    ret.p = 1, ret.d = 0;
    while(b){
        if(b & 1) ret = mul_er(ret, a, mod);
        a = mul_er(a, a, mod);
        b >>= 1;
    }
    return ret;
}
ll legendre(ll a, ll p){
    return ppow(a, (p - 1) >> 1, p);
}
ll modulo(ll a, ll mod){
    a %= mod;
    if(a < 0) a += mod;
    return a;
}
ll solve(ll n, ll p){   //x^2 = n mod p
    if(n == 0) return 0;
    if(n == 1) return 1;
    if(p == 2) return 1;
    if(legendre(n, p) + 1 == p) return -1;  //无解
    ll a = -1, t;
    while(true){
        a = rand() % p;
        t = a * a - n;
        w = modulo(t, p);
        if(legendre(w, p) + 1 == p) break;
    }
    TT temp;
    temp.p = a;
    temp.d = 1;
    TT ans = power(temp, (p + 1) >> 1, p);
    return ans.p;
}

int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        ll a, n;
        scanf("%lld%lld", &a, &n);
        ll ans1 = solve(a, n), ans2;
        if(ans1 == -1){
            printf("No root\n");
            continue;
        }
        ans2 = n - ans1;
        if(ans1 > ans2) swap(ans1, ans2);
        if(ans1 == ans2) printf("%lld\n", ans1);
        else printf("%lld %lld\n", ans1, ans2);
    }
    return 0;
}