【简单】LightOJ Integer Divisibility 1078

1078 - Integer Divisibility

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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 of that number in decimal notation is a sequence of only a digit. Now you are given the number and the only allowable digit, you should report the number of digits of such multiple.

For example you have to find a multiple of 3 which contains only 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 of digits of such multiple. If several solutions are there; report the minimum one.

Sample Input

Output for Sample Input

3

3 1

7 3

9901 1

Case 1: 3

Case 2: 6

Case 3: 12

 

题意：

给一个数n，再给一个digit，问digit组成的多少位的数可以整除n。

AC代码：

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int main(){    int t;    int xp=1;    scanf("%d",&t);    while(t--){        long long n,digit;        scanf("%lld%lld",&n,&digit);        long long X=0;        int num=0;        while(1){            X=X*10+digit;            num++;            //if(X%2==0||X%5==0)continue;            if(X%n==0)break;            X%=n;        }        printf("Case %d: %lld\n",xp++,num);    }}