LightOJ 1078 - Integer Divisibility (同余定理)

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 andn will not be divisible by2 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：给出两个数n和digit，求要几个（cnt）digit能把n整除。  例如给出3和1,   111能整除3， 所以cnt为3。

题解：很明显的同余定理，(a+b)%n=(a%n+b%n)%n。

代码如下：

#include<cstdio>#include<cstring>int main(){int n,digit,ans,cnt;int k=1,t;scanf("%d",&t);while(t--){scanf("%d%d",&n,&digit);ans=digit%n;//赋值前要先取余，要不然会WA cnt=1;while(ans){ans=(ans*10+digit)%n;cnt++;}printf("Case %d: %d\n",k++,cnt);}return 0;}