Hello Kiki （中国剩余定理模版）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3579 

#include <iostream>
#include <algorithm>

typedef long long LL;

using namespace std;




LL ex_gcd(LL a,LL b,LL &x,LL &y){
    if (b == 0){
        x = 1;
        y = 0;
        return a;
    }
    LL gcd = ex_gcd(b,a%b,y,x);
    y -= (a/b)*x;
    return gcd;
}

LL inv(LL t,LL p){   // t 关于 p 的逆元
    LL x,y,d;
    d = ex_gcd(t,p,x,y);
    if (d == 1){
        return (x%p+p)%p;
    }else
        return -1;
}


// n个方程： x = a[i](mod m[i])   (0<=i<n)

LL china(int n,LL *a,LL *m){
    LL M = 1,ret = 0;
    for (int i=0;i<n;i++){
        M *= m[i];
    }
    for (int i=0;i<n;i++){
        LL w = M/m[i];
        ret = (ret + inv(w,m[i]) * w * a[i]) % M;
    }
    return (ret + M) % M;
}

LL CRT(LL b[],LL n[],int num){
    bool flag = false;
    LL n1 = n[0],n2,b1 = b[0],b2,bb,d,t,k,x,y;
    for (int i=1;i<num;i++){
        n2 = n[i],b2 = b[i];
        bb = b2 - b1;
        d = ex_gcd(n1,n2,x,y);
        if (bb%d){
            flag = true;
            break;
        }
        k = bb / d * x;
        t = n2 / d;
        if (t<0)
            t = -t;
        k = (k % t + t) % t;
        b1 = b1 + n1 * k;
        n1 = n1 / d * n2;
    }
    if (flag)
        return -1;
    if (b1 == 0){
        b1 = n1;
    }
    return b1;
}


int main(){
    int t,num;
    LL b[55],n[55];
    scanf("%d",&t);
    int cas = 1;
    while (t--){
        scanf("%d",&num);
        for (int i=0;i<num;i++){
            scanf("%lld",&n[i]);
        }
        for (int i=0;i<num;i++){
            scanf("%lld",&b[i]);
        }
        printf("Case %d: %lld
",cas++,CRT(b,n,num));
    }
    return 0;
}