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
",i,ans);
    }
    return 0;
}