数论-扩展中国剩余定理

求解 n 个同余方程

x ≡ Ci (mod Mi)

推荐一下讲证明的文章：

1.https://www.cnblogs.com/TheRoadToTheGold/p/8458326.html 很简洁易懂。

2.http://blog.csdn.net/clove_unique/article/details/54571216 排版不太好；这篇文章还讲了扩展卢卡斯 (Lucas) 定理。

 可以用滚动数组优化空间，但我的代码里没这么写。

void China()
{
    scanf("%lld", &n);
    for (i = 1; i <= n; ++i)
        scanf("%lld%lld", &M[i], &C[i]);
    bool flag = false;
    for (i = 2; i <= n; ++i) {
        LL m1 = M[i-1], m2 = M[i], c1 = C[i-1], c2 = C[i];
        LL t = gcd(m1, m2);
        if ((c2-c1) % t) { flag = true; break; }
        M[i] = m1*m2 / t;
        C[i] = Inv(m1/t, m2/t) * (c2-c1)/t % (m2/t) * m1 + c1;
        C[i] = (C[i]%M[i]+M[i]) % M[i];
    }
    if (flag) { printf("-1
"); continue; }
    printf("%lld
", C[n]);
}

模板题 POJ2891

#include <stdio.h>

typedef long long LL;

const int _N = 1200;

LL C[_N], M[_N];

void exgcd(LL m, LL n, LL &x, LL &y, LL &d)
{
    if (n) { exgcd(n, m%n, y, x, d); y -= m/n*x; return; }
    x = 1, y = 0, d = 1;
    return;
}

LL gcd(LL t1, LL t2) { return t2 ? gcd(t2, t1 % t2) : t1; }

LL Inv(LL v, LL p)
{
    LL x, y, d;
    exgcd(v, p, x, y, d);
    x = (x%p+p)%p;
    if (!x) x = p;
    return x;
}

int main()
{
    LL n, i;
    while (~scanf("%lld", &n)) {
        for (i = 1; i <= n; ++i)
            scanf("%lld%lld", &M[i], &C[i]);
        bool flag = false;
        for (i = 2; i <= n; ++i) {
            LL m1 = M[i-1], m2 = M[i], c1 = C[i-1], c2 = C[i];
            LL t = gcd(m1, m2);
            if ((c2-c1) % t) { flag = true; break; }
            M[i] = m1*m2 / t;
            C[i] = Inv(m1/t, m2/t) * (c2-c1)/t % (m2/t) * m1 + c1;
            C[i] = (C[i]%M[i]+M[i]) % M[i];
        }
        if (flag) { printf("-1
"); continue; }
        printf("%lld
", C[n]);
    }
    return 0;
}