#833010 (C++17) No.2191 一元二次式 mod 奇素数

• 問題: No.2191 一元二次式 mod 奇素数
• ユーザー: milanis48663220
• 提出日時: 2023-01-13 23:09:34
• 言語: C++17 (gcc 13.3.0 + boost 1.87.0)
• 結果: AC
• 実行時間: 22 ms / 2,000 ms
• コード長: 2,554 bytes
• コンパイル時間: 893 ms
• コンパイル使用メモリ: 116,416 KB
• 最終ジャッジ日時: 2025-02-10 03:12:33
• ジャッジサーバーID （参考情報）: judge1 / judge4

ファイルパターン	結果
other	AC * 27

ソースコード

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#include <iostream>
#include <algorithm>
#include <iomanip>
#include <vector>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <tuple>
#include <cmath>
#include <numeric>
#include <functional>
#include <cassert>
#include <atcoder/modint>
#define debug_value(x) cerr << "line" << __LINE__ << ":<" << __func__ << ">:" << #x << "=" << x << endl;
#define debug(x) cerr << "line" << __LINE__ << ":<" << __func__ << ">:" << x << endl;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
using namespace std;
typedef long long ll;
template<typename T>
vector<vector<T>> vec2d(int n, int m, T v){
    return vector<vector<T>>(n, vector<T>(m, v));
}
template<typename T>
vector<vector<vector<T>>> vec3d(int n, int m, int k, T v){
    return vector<vector<vector<T>>>(n, vector<vector<T>>(m, vector<T>(k, v)));
}
template<typename T>
void print_vector(vector<T> v, char delimiter=' '){
    if(v.empty()) {
        cout << endl;
        return;
    }
    for(int i = 0; i+1 < v.size(); i++) cout << v[i] << delimiter;
    cout << v.back() << endl;
}
// from https://judge.yosupo.jp/submission/84154
using u64 = unsigned long long;
using u32 = unsigned;
u32 modpow(u32 a, u32 n, u32 p){
  u32 r = 1;
  for(; n; n >>= 1){
    if(n & 1) r = (u64) r * a % p;
    a = (u64) a * a % p;
  }
  return r;
}
bool is_square(u32 n, u32 p){
  n %= p;
  return n == 0 || modpow(n, (p-1)/2, p) == 1;
}
u32 sqrt_mod(u32 n, u32 p){
  u32 a = 2;
  while(is_square(((u64) a * a + p - n) % p, p)){
    a++;
  }
  a = 2 * a % p;
  u64 s = 1;
  u64 t = p - a;
  u64 u = p - a;
  u64 v = n;
  u32 m = (p+1)/2;
  while(m) {
    if(m & 1){
      s = ((p - s) * u + t) % p;
      t =  t * v % p;
    }
    else {
      t = (s * v + (p - t) * u) % p;
    }
    u = (2 * v + (p - u) * u) % p;
    v = v * v % p;
    m /= 2;
  }
  return s;
}
using mint = atcoder::modint;
ostream& operator<<(ostream& os, const mint& m){
    os << m.val();
    return os;
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << setprecision(10) << fixed;
    int p; cin >> p;
    mint::set_mod(p);
    mint k = (mint(p)-1)/2;
    mint x = k*k+4*k+2;
    mint s = mint(9)-4*x;
    if(!is_square(s.val(), p)){
        cout << "NO" << endl;
        return 0;
    }
    cout << "YES" << endl;
    // auto ans = mint(sqrt_mod(s.val(), p));
    // if(ans*ans == s){
    // }else{
    //     cout << "NO" << endl;
    // }
}
 
 
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