Light Oj 1078 Integer Divisibility(同余定理）

Integer Divisibility

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

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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

思路：中间对d累加的过程中，会超long long ,所以每次都先取余，再累加；这是同余定理的应用；

代码：

#include<stdio.h>int main(){int t,n,d,i,mm=1;scanf("%d",&t);while(t--){    scanf("%d%d",&n,&d);long long sum=d;for(i=1;;i++){sum=sum%n;if(sum==0)break;sum=sum*10+d;}printf("Case %d: %d\n",mm++,i);}return 0;}