整除-- 同余定理

最新推荐文章于 2022-05-08 15:25:35 发布

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

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 1

7 3

9901 1

Sample Output

Case 1: 3

Case 2: 6

Case 3: 12

解题思路：给你两个整数n和dight，其中n不能被2和5整除，0<dight<10，问有多少位dight能整除n。代码写起来很简单，刚开始用了个for循环，一提交就超时。。。

因为数据有限制，所以精简代码，用同余定理！（a+b）%c=（a%c+b%c）%c，乘法也一样。

#include<cstdio>
int main(){
    int n,a,t,ans=1;
    scanf("%d",&t);
    while(t--){
    	int count=1;//加上个位数，初始化为1 
    	scanf("%d %d",&n,&a);
    	int m=a%n;//赋值时就要取余，不然会WA 
    	while(m){
    		m=m*10+a;
    		count++;
    		m=m%n;//再次取余 
		}
    	printf("Case %d: %d\n",ans++,count);
	}
    return 0;  
}