【light oj1078】Integer Divisibility（同餘定理）

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Time Limit: 2 second(s)

Memory Limit: 32 MB

If an integer is not divisible by 2 or 5, some multiple ofthat number in decimal notation is a sequence of only a digit. Now you aregiven the number and the only allowable digit, you should report the number ofdigits of such multiple.

For example you have to find a multiple of 3 which containsonly 1's. Then the result is 3 because is 111 (3-digit) divisible by 3.Similarly if you are finding some multiple of 7 which contains only 3's then,the result is 6, because 333333 is divisible by 7.

Input

Input starts with an integer T (≤ 300),denoting the number of test cases.

Each case will contain two integers n (0 < n ≤10⁶ and n will not be divisible by 2 or 5)and the allowable digit (1 ≤ digit ≤ 9).

Output

For each case, print the case number and the number ofdigits of such multiple. If several solutions are there; report the minimumone.

Sample Input	Output for Sample Input
3 3 1 7 3 9901 1	Case 1: 3 Case 2: 6 Case 3: 12

考察同餘定理，因爲數據過大可能會超long long 型。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#define LL long long
#include<algorithm>
using namespace std;
int main(){
	int T,cases=0;
	scanf("%d",&T);
	while(T--){
		LL n,m;
		scanf("%lld%lld",&n,&m);
		LL ans=1,cnt=m;
		while(cnt%n!=0){
			cnt=(cnt*10+m)%n;
			ans++;
		}
		printf("Case %d: %lld\n",++cases,ans);
	}
	return 0;
}