light oj 1078 - Integer Divisibility

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 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 n和m，让判断多少位的m可以整除n（每一位的数值都为m，如果当前的位数不能整除n则再加上一位）如：3   1意思是多少个1可以整除3显然111可以整除3，所以就是三位，输出3

小技巧：令ans=m 让ans对n取余为ans（覆盖原来的值），每次让ans=ans*10+m（因为不能整除所以增加位数）直到可以整除   原理大概是除法的同余定理吧

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#include<math.h>
#define LL long long
#define DD double
#define MAX 10000
using namespace std;
int main()
{
    int t;
    int n,m,j,i;
    LL ans;
    scanf("%d",&t);
    int k=1;
    while(t--)
    {
        scanf("%d%d",&n,&m);
        printf("Case %d: ",k++);
        ans=m;
        int sum=1;
        while(ans%n)
        {
            ans=ans%n;
            ans=ans*10+m;
            sum++;
        }
        printf("%d\n",sum);
    }
    return 0;
}