河南多校大一训练赛 D

题目链接：http://acm.hust.edu.cn/vjudge/contest/125004#problem/D

密码：acm

Description

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

Sample Output
Case :
Case :
Case : 

分析：

第一次代码，超了应该，反正看输不出结果：

 #include<cstdlib>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 using namespace std;

 int main()
 {
     int n,i,o=,a,b;
     scanf("%d", &n);

     for(i=;i<=n;i++)
     {
         scanf("%d%d", &a,&b);
         int d=b,ans=;

         while(d%a)
         {
             d=d*+b;
             ans++;
         }

         printf("Case %d: %d\n", o++,ans);
     }
     return ;
 }

修改后：

 #include<cstdlib>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 using namespace std;

 int main()
 {
     int n,i,o=,a,b;
     scanf("%d", &n);

     for(i=;i<=n;i++)
     {
         scanf("%d%d", &a,&b);
         int d=b,ans=,r;

         r=b%a;
         while(r)
         {
             d=r*+b;
             ans++;
             r=d%a;
         }

         printf("Case %d: %d\n", o++,ans);
     }
     return ;
 }