Light 1078 Integer Divisibility (取模运算)

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 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

因为到最后数会很大，取模才行

代码：

#include<cstdio>int main(){int ans,t,n,m,d,k=0;scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);ans=1;if(m%n==0){printf("Case %d: 1\n",++k);continue;}else{d=m;while(d%n){d=d%n;d=d*10+m;ans++;}printf("Case %d: %d\n",++k,ans);}}return 0;}