Light 1078 Integer Divisibility (取模运算)

      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 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 andn 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

大意：就是十进制的数n，至少给定多少n能被整除

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#define LL long longusing 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;}