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  • 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
    ", flag ? C[K] : -1);
        }
        return 0;
    }

    参考链接:https://www.cnblogs.com/zwfymqz/p/8425731.html

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  • 原文地址:https://www.cnblogs.com/lfri/p/11372164.html
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