扩展中国剩余定理【模板】

其实扩展中国剩余定理与中国剩余定理的关系不大，后者使用类似拉格朗日插值法构造解，前者是模方程两两合并。

C++版

#include<iostream>
#include<cstdio>
#define LL long long //or __int128
using namespace std;
const LL MAXN = 1e6 + 10;
LL K, C[MAXN], M[MAXN], x, y;
LL gcd(LL a, LL b) {
    return b == 0 ? a : gcd(b, a % b);
}
LL exgcd(LL a, LL b, LL &x, LL &y) {
    if (b == 0) {x = 1, y = 0; return a;}
    LL r = exgcd(b, a % b, x, y), tmp;
    tmp = x; x = y; y = tmp - (a / b) * y;
    return r;
}
LL inv(LL a, LL b) {
    LL r = exgcd(a, b, x, y);
    while (x < 0) x += b;
    return x;
}
int main() {
    while (~scanf("%lld", &K)) {
        for (LL i = 1; i <= K; i++) scanf("%lld%lld", &M[i], &C[i]); //x = C[i](mod M[i])
        bool flag = 1;
        for (LL i = 2; i <= K; i++) {
            LL M1 = M[i - 1], M2 = M[i], C2 = C[i], C1 = C[i - 1], T = gcd(M1, M2);
            if ((C2 - C1) % T != 0) {flag = 0; break;}
            M[i] = (M1 * M2) / T;  //可能爆long long
            C[i] = ( inv( M1 / T , M2 / T ) * (C2 - C1) / T ) % (M2 / T) * M1 + C1;
            C[i] = (C[i] % M[i] + M[i]) % M[i];
        }
        printf("%lld\n", flag ? C[K] : -1);
    }
    return 0;
}

 

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个性签名：时间会解决一切