模板——中国剩余定理

 1 //n个方程，x=a[i](mod m[i]) 
 2 
 3 LL china(int n, int  *a, int *m) {
 4     LL M = 1, d, y, x = 0;
 5     for(int i = 0; i < n; i++) M *= m[i];
 6     for(int i = 0; i < n‘ i++) {
 7         LL w = M / m[i];
 8         exgcd(m[i], w, d, d, y);
 9         x = (x + y*w*a[i]) % M;
10     }
11     return (x + M) % M;
12 }