模板——中国剩余定理

//n个方程，x=a[i](mod m[i]) 

LL china(int n, int  *a, int *m) {
    LL M = 1, d, y, x = 0;
    for(int i = 0; i < n; i++) M *= m[i];
    for(int i = 0; i < n' i++) {
        LL w = M / m[i];
        exgcd(m[i], w, d, d, y);
        x = (x + y*w*a[i]) % M;
    }
    return (x + M) % M;
}