HOJ 1008 How many N

最新推荐文章于 2019-06-25 22:47:32 发布

DyanWang

最新推荐文章于 2019-06-25 22:47:32 发布

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分类专栏： HOJ 文章标签： c C hoj HOJ

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此博客探讨如何找到一个由N位数字构成的最小整数K，该整数可以被M整除。通过输入一对正整数N和M，输出满足条件的K，或若不存在输出零。

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Time limit : 15 sec

Memory limit : 32 M

Find a minimal interger K which is merely comprised of N and can be divided by M.

For example,11 is the minimal number that and be divided by 11, and it is comprised of two '1's, and 111111 can be divided by 13 which is comprised of six '1's.

Input

On each line of input , there will be two positive integer, N and M. N is a digit number, M is no more than 10000.

Output

On each single line, output the number of N, if no such K, output zero.

Sample Input

1 5
1 11
1 13

Sample Output

0
2
6

Solution:

int main( )
{
    int n, m, t, f[ 10000 ], i, count;
    while ( scanf("%d %d", &n, &m) == 2 )
    {
        count = t = 0;
        for ( i = 0; i < m; i++ )
            f[ i ] = 0;
        while ( !f[ t ] )
        {
            count++;
            f[ t ] = 1;
            t = ( t * 10 + n ) % m;
        }
       if ( !t )
            printf("%d\n", count);
       else
            printf("0\n");
    }
    return 0;
}