#20797 (C++11（廃止可能性あり）) No.186 中華風 (Easy)

• 問題: No.186 中華風 (Easy)
• ユーザー: Lepton_s
• 提出日時: 2015-04-20 00:14:44
• 言語: C++11（廃止可能性あり） (gcc 13.3.0)
• 結果: AC
• 実行時間: 2 ms / 2,000 ms
• コード長: 1,609 bytes
• コンパイル時間: 815 ms
• コンパイル使用メモリ: 87,564 KB
• 実行使用メモリ: 5,376 KB
• 最終ジャッジ日時: 2024-07-19 18:22:54
• 合計ジャッジ時間: 1,654 ms
• ジャッジサーバーID （参考情報）: judge1 / judge5

ファイルパターン	結果
other	AC * 23

ソースコード

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#include <algorithm>
#include <climits>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <sstream>
#include <functional>
#include <map>
#include <string>
#include <cstring>
#include <vector>
#include <queue>
#include <stack>
#include <deque>
#include <set>
#include <list>
#include <numeric>
using namespace std;
const double PI = 3.14159265358979323846;
const double EPS = 1e-12;
const int INF = 1<<25;
typedef pair<int,int> P;
typedef __int128 ll;
typedef unsigned long long ull;
ll mod = 1e9+7;
ll pow_mod(ll a, ll n, ll m){
    ll r = 1;
    while(n){
        if(n&1) r = r*a%m;
        a = a*a%m;
        n >>= 1;
    }
    return r;
}
ll extgcd(ll a, ll b, ll &x, ll &y){
    ll d = a;
    if(b){
        d = extgcd(b, a%b, y, x);
        y -= (a/b)*x;
    } else {
        x = 1; y = 0;
    }
    return d;
}
ll gcd(ll a, ll b){
    ll x, y;
    return extgcd(a, b, x, y);
}
ll mod_inv(ll a, ll m){
    ll x, y;
    extgcd(a, m, x, y);
    return (m+x%m)%m;
}
int main(){
    ll n;
    // cin>>n;
    n = 3;
    vector<ll> b(n), m(n);
    for(int i = 0; i < n; i++){
        int bb, mm;
        cin>>bb>>mm;
        b[i] = bb;
        m[i] = mm;
    }
    ll res = -2;
    /*
    for(int i = 0; i < n; i++){
        for(int j = i+1; j < n; j++){
            ll gm = __gcd(m[i], m[j]);
            if(b[i]%gm!=b[j]%gm) res = -1;
        }
    }
    if(res==-1){
        cout<<-1<<endl;
        return 0;
    }
    */
    ll X = 0, M = 1;
    for(int i = 0; i < n; i++){
        ll A = M, B = b[i]-X, D = gcd(m[i], M);
        if(B%D){
            res = -1;
            break;
        }
        A /= D;
        B /= D;
        m[i] /= D;
        ll T = B%m[i]*mod_inv(A, m[i])%(m[i]);
        X = X+M*T;
        M *= m[i];
        X = (X%M+M)%M;
    }
    cout<<(long long)((res==-1)?-1:(X-1+M)%M+1)<<endl;
    return 0;
}
 
 
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