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.

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define LL long long
using namespace std;

int main()
{
    LL n,m,i;
    LL cla;
    scanf("%lld",&cla);
    for(int gr=1;gr<=cla;gr++)
    {
        scanf("%lld%lld",&n,&m);
        printf("Case %lld: ",gr);
        LL ans=1;
        LL t=m;
        while(t%n!=0)
        { 
            t=(t*10+m)%n;//如果不取余n数据大会TLE
            ans++;
        }
        printf("%lld\n",ans);
    }
    return 0;
}

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