Good Number

Time Limit: 1000ms

Problem Description:

       Let's call a number k-good if it contains all digits not exceeding k (0, ..., k). You've got a number k and an array a containing n numbers. Find out how many k-good numbers are in a (count each number every time it occurs in array a).

Input:

The input includes several cases. For each case, the input format is:
The first line contains integers n and k (1 ≤ n ≤ 100, 0 ≤ k ≤ 9). The i-th of the following n lines contains integer ai without leading zeroes (1 ≤ ai ≤ 109).

Output:

For each case, Print a single integer — the number of k-good numbers in a.

Sample Input:

10 6
1234560
1234560
1234560
1234560
1234560
1234560
1234560
1234560
1234560
1234560
2 1
1
10
2 0
10
20

Sample Output:

10
1
2

感触：做了很久，没有ac不了的水题！！！！

import java.util.Arrays;
import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        int i,j;
        Scanner cin = new Scanner(System.in);
        while(cin.hasNext())
        {
            int num = 0;
            int n = cin.nextInt();            //测试数据
            int m = cin.nextInt();             //不能超过的数
            while(n!=0)
            {
                int x = cin.nextInt();
                String st = Integer.toString(x, 10); //转换成字符
                int a = st.length();
                boolean  h[] = new boolean[m+1];
                for(j=0;j<a;j++)       //检查这个数是否包含从0到m的数
                {
                    for(i=0;i<=m;i++)   //
                    {
                        int t = ((int)st.charAt(j)-(int)('0'));
                        if( t== i)  //判断每一位是否都大于那个数
                            h[i] = true;
                    }
                }
                int flag =1;
                for(i=0;i<=m;i++)
                {
                    if(h[i]!=true)
                        flag =0;
                }
                if(flag==1)num++;
                n--;

            }
            System.out.println(num);
        }
    }
}