偷偷放個大佬的教學:中國剩餘定理 互質的情況

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

不互質的情況

ll ex_crt(int n, int *a, int *m){
	if(n == 1 &&
 
 a[1] == 0) return m[1];
    ll M = m[1],A = a[1], t, d, x, y;
    for(int i = 2; i <= n; i++){
        d = exgcd(M,m[i],x,y);
        if((a[i]-A)%d)return -1;
        x *= (a[i]-A)/d;
        t = m[i]/d; x = (x%t+t)%t;
        A = M*x+A; M = M/d*m[i];
        A%=M;
    }
    return (A+M)%M;
}