HDU 3579 Hello Kiki中国の残りの定理

Hello Kiki
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 6297 Accepted Submission(s): 2404
Problem Description One day I was shopping in the supermarket.There was a cashier counting coins seriously when a little kid running and singing「門前大橋の下流にアヒルの群れがいるから、早く数えなさい.二四六七八」. And then the cashier put the counted coins back morosely and count again… Hello Kiki is such a lovely girl that she loves doing counting in a different way. For example, when she is counting X coins, she count them N times. Each time she divide the coins into several same sized groups and write down the group size Mi and the number of the remaining coins Ai on her note. One day Kiki’s father found her note and he wanted to know how much coins Kiki was counting.
Input The first line is T indicating the number of test cases. Each case contains N on the first line, Mi(1 <= i <= N) on the second line, and corresponding Ai(1 <= i <= N) on the third line. All numbers in the input and output are integers. 1 <= T <= 100, 1 <= N <= 6, 1 <= Mi <= 50, 0 <= Ai < Mi
Output For each case output the least positive integer X which Kiki was counting in the sample output format. If there is no solution then output -1.
Sample Input 2 2 14 57 5 56 5 19 54 40 24 80 11 2 36 20 76
Sample Output Case 1: 341 Case 2: 5996
問題の接続
問題の説明
違う方法で数えるのが好きな女の子がいて、今彼女はN種類の方法でX個(未知)の硬貨を数えて、Mi個の硬貨は1組の数で、Ai個を残して、iは1からNです.これらのコインは最低何個あるかを聞いて、答えがあればXを出力して、さもなくば-1を出力します.
もんだいぶんせき
明らかに中国の残りの定理で解き、テンプレート問題である.Mi両両は必ずしも互いに質的ではないため、原版の中国の残りの定理で解くことができず、拡張版を使う.一つ注意すべきことは、型抜き後の結果が0の場合、Mi(iは1からn)の最小公倍数を出力することである.
コードは次のとおりです.

#include
using namespace std;
typedef long long ll;
const int N=10; 

int n;
ll m[N],a[N];

ll exgcd(ll a,ll b,ll &x,ll &y){
	if(b==0){
		x=1;
		y=0;
		return a;
	}
	ll ans,x1,y1;
	ans=exgcd(b,a%b,x1,y1);
	x=y1;
	y=x1-(a/b)*y1;
	return ans;
}

ll CRT(){
	ll M,R,x,y,g,c;
	R=a[0];
	M=m[0];
	for(int i=1;i<n;i++){
		g=exgcd(M,m[i],x,y);
		c=a[i]-R;
		if(c%g) return -1;
		ll t=m[i]/g;
		x=(c/g*x)%t;
		R=M*x+R;
		M=M/g*m[i];
		R%=M;
	}
	R=(R+M)%M;
	if(R==0) R=M;
	return R;
}

int main(){
	ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
	int time;
	cin>>time;
	for(int t=1;t<=time;t++){
		cin>>n;
		for(int i=0;i<n;i++) cin>>m[i];
		for(int i=0;i<n;i++) cin>>a[i];
		cout<<"Case "<<t<<": "<<CRT()<<endl;
	}
	return 0;
}