Light OJ Integer Divisibility 【取模】

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

<span style="font-family:Courier New;font-size:12px;">#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long LL;
char s[10010];
int main()
{
int t,n,num=0;
LL d;
scanf("%d",&t);
while(t--)
{
scanf("%d%lld",&n,&d);
printf("Case %d: ",++num);
LL tem=d;
int cnt=1;
while(d%n)
{
d=(d*10+tem)%n;//这里不取余会超时
cnt++;
}
printf("%d\n",cnt);
}
return 0;
}</span>