結果

問題 No.3022 一元一次式 mod 1000000000
ユーザー 👑 binap
提出日時 2025-02-14 21:46:18
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 294 ms / 2,000 ms
コード長 3,054 bytes
コンパイル時間 3,379 ms
コンパイル使用メモリ 249,736 KB
実行使用メモリ 6,820 KB
最終ジャッジ日時 2025-02-17 12:56:28
合計ジャッジ時間 4,991 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
other AC * 21
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ソースコード

diff #
プレゼンテーションモードにする

#include<bits/stdc++.h>
#include<atcoder/all>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
using namespace atcoder;
typedef long long ll;
typedef vector<int> vi;
typedef vector<long long> vl;
typedef vector<vector<int>> vvi;
typedef vector<vector<long long>> vvl;
typedef long double ld;
typedef pair<int, int> P;
template <int m> ostream& operator<<(ostream& os, const static_modint<m>& a) {os << a.val(); return os;}
template <int m> ostream& operator<<(ostream& os, const dynamic_modint<m>& a) {os << a.val(); return os;}
template <int m> istream& operator>>(istream& is, static_modint<m>& a) {long long x; is >> x; a = x; return is;}
template <int m> istream& operator>>(istream& is, dynamic_modint<m>& a) {long long x; is >> x; a = x; return is;}
template<typename T> istream& operator>>(istream& is, vector<T>& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;}
template<typename U, typename T> ostream& operator<<(ostream& os, const pair<U, T>& p){os << p.first << ' ' << p.second; return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<T>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return 
    os;}
template<typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : ""); 
    return os;}
template<typename T> ostream& operator<<(ostream& os, const set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename T> ostream& operator<<(ostream& os, const unordered_set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename S, auto op, auto e> ostream& operator<<(ostream& os, const atcoder::segtree<S, op, e>& seg){int n = seg.max_right(0, [](S){return 
    true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}
template<typename S, auto op, auto e, typename F, auto mapping, auto composition, auto id> ostream& operator<<(ostream& os, const atcoder
    ::lazy_segtree<S, op, e, F, mapping, composition, id>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 
    1 ? "\n" : " "); return os;}
template<typename T> void chmin(T& a, T b){a = min(a, b);}
template<typename T> void chmax(T& a, T b){a = max(a, b);}
template<typename T>
T pow_mod(T A, T N, T MOD){
    T res = 1 % MOD;
    A %= MOD;
    while(N){
        if(N & 1) res = (res * A) % MOD;
        A = (A * A) % MOD;
        N >>= 1;
    }
    return res;
}
const long long e9 = 1000000000;
long long solve(long long n, long long m, long long MOD){
    long long g = gcd(MOD, n);
    if(g >= 2){
        if(m % g == 0) return solve(n / g, m /g, MOD / g);
        else return -1LL;
    }
    long long phi = MOD;
    if(MOD % 2 == 0) phi /= 2;
    if(MOD % 5 == 0){
        phi /= 5;
        phi *= 4;
    }
    long long n_inv = pow_mod(n, phi - 1, MOD);
    long long ans = (((-m % MOD) + MOD) * n_inv) % MOD;
    if(ans == 0) ans += MOD;
    return ans;
}
int solve(){
    long long n, m;
    cin >> n >> m;
    
    cout << solve(n, m, e9) << "\n";
    return 0;
};
int main(){
    int t;
    cin >> t;
    rep(_, t) solve();
    return 0;
}
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