中国剩余定理 [mi不互质]

假设有如下方程: 

合并可得: 

然后利用扩展欧几里得算法, 算出x1的最小正整数解, 代入原方程即可得y的值, 然后求出最终答案

应用: http://blog.csdn.net/hnust_xx/article/details/76130227

struct P {  
    ll as, lc; ///as为ai, lc为最小公倍数
    P() {}
    P(ll as, ll lc) : as(as), lc(lc) {}  
};  

ll exgcd(ll a, ll b, ll &x, ll &y) {  
    if(b == 0) {  
        x = 1; y = 0;  
        return a;  
    }  
    ll ans = exgcd(b, a % b, x, y);  
    ll t = x;  
    x = y;  
    y = t - a / b * y;  
    return ans;  
}  

ll inv(ll t, ll p) {
    ll d, x, y;
    d = exgcd(t, p, x, y);
    return (x % p + p) % p;
}

P Merge(P gg, P rr) {
    if(gg.as == -1 || rr.as == -1) return P(-1, -1);
    ll a1 = gg.as, b1 = gg.lc;
    ll a2 = rr.as, b2 = rr.lc;
    ll kx, ky;
    ll g = exgcd(b1, b2, kx, ky);
    if((a2 - a1) % g) return P(-1, -1);
    ll p = ((a2 - a1) % b2 + b2) % b2;
    p /= g;
    p *= inv(b1 / g, b2 / g);
    p %= b2;
    p *= b1;
    p += a1;
    ll b3 = b1 / g * b2;
    ll a3 = (p % b3 + b3) % b3;
    return P(a3, b3);
}

ll CRT(ll *a, ll *m, int n) {
    P g(a[1], m[1]);
    for(int i = 1; i < n; i++) {
        g = Merge(g, P(a[i], m[i]));
        if(g.as == -1) return g;
    }
    return g;
}