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
3
3 1
7 3
9901 1
Sample Output
Case 1: 3
Case 2: 6
Case 3: 12

 

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=1,a,b;
    scanf("%d", &n);

    for(i=1;i<=n;i++)
    {
        scanf("%d%d", &a,&b);
        int d=b,ans=1;

        while(d%a)
        {
            d=d*10+b;
            ans++;
        }

        printf("Case %d: %d\n", o++,ans);
    }
    return 0;
}

 

 

修改后：

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

int main()
{
    int n,i,o=1,a,b;
    scanf("%d", &n);

    for(i=1;i<=n;i++)
    {
        scanf("%d%d", &a,&b);
        int d=b,ans=1,r;

        r=b%a;
        while(r)
        {
            d=r*10+b;
            ans++;
            r=d%a;
        }

        printf("Case %d: %d\n", o++,ans);
    }
    return 0;
}