light oj 1078 - Integer Divisibility (取模运算)

1078 - Integer Divisibility

   PDF (English)StatisticsForum

Time Limit: 2 second(s)Memory Limit: 32 MB

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

Output for Sample Input

3

3 1

7 3

9901 1

Case 1: 3

Case 2: 6

Case 3: 12

 题意：给出一个数n 和 s, 问只由s组成的数多少位时可以被n整除。

思路：这题整除，每次增加一位s,当余数等0的时候就是所求。

代码：

#include <iostream>#include <cstdio>using namespace std;int main(){    int t,kcase=1;    int n,s;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&s);        int ans,x,a,b;        ans=1;        x=s;        while(x%n!=0)        {            x=(x*10+s)%n;            ans++;        }        printf("Case %d: %d\n",kcase++,ans);    }    return 0;}