LightOj 1078 Basic Math

思路：

设输入的两个数分别为n和a,每一次所得到的数为update：

开始update=a,依次update分别为update*10+a,这样数据会超出范围，则update每次为update=(update*10+a)%n即可，

如果update=0,跳出循环；

只需证明：(update*10+a)%n=(update%n*10+a)%n即可；

由(update*10+a)%n=(update%n*10+a%n)%n，因为a<=n，所以a%n=a.证必；

1078 - Integer Divisibility

PDF (English) Statistics Forum

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

PROBLEM SETTER: JANE ALAM JAN

/********************************
author   : Grant Yuan
time     : 2014/8/21 0:28
algorithm: Basic Math
source   : LightOj 1078
**********************************/
#include<bits/stdc++.h>

using namespace std;
int t;
int a,b,ans;
int main()
{
scanf("%d",&t);
for(int i=1;i<=t;i++)
{
ans=1;
scanf("%d%d",&a,&b);
int temp=b;
while(temp%a!=0)
{
temp=temp*10;
temp+=b;
temp%=a;
ans++;
}
printf("Case %d: %d\n",i,ans);
}
return 0;
}