| 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
",sum);
}
return 0;
}