2013暑期第一周周赛G题 错误总结

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 first line contains integers n and k (1 ≤ n ≤ 100, 0 ≤ k ≤ 9). The i-th of the following n lines contains integer a_i without leading zeroes (1 ≤ a_i ≤ 10⁹).

Output

Print a single integer — the number of k-good numbers in a.

Sample Input

Input

10 6

1234560

1234560

1234560

1234560

1234560

1234560

1234560

1234560

1234560

1234560

Output

10

Input

2 1

1

10

Output

1

代码

#include<stdio.h>

#include<string.h>

int main()

{

	char s[105][105], k;

	int n, a;

	int i, j;

	scanf("%d %c", &n, &k);

	getchar(); //这个地方，要加getchar();

	for (i = 0; i < n; i++)

	{

		gets(s[i]);//他的后面不需getchar();

	}

	a = k - '0';

	//printf("a = %d\n",a);

	int len, sum = 0, x = 0, bleg = 0, t = 0;

	for (i = 0; i < n; i++)

	{

		int f[11] = {0};//这个地方，标记数组必须要放到函数体内部，开始放在了外部，全局变量，每当第二次输入，若第二次长度小于第一次，则不会清零；

		bleg = 0; //bleg需要在这里标记，不能放到外面

		len = strlen(s[i]);		//printf("len s[%d] = %d\n",i,len);

		if (len < a + 1)

		{

			bleg = 0;

		}

		else

		{

			t = 0;

			for (j = 0; j < len; j++)

			{

				int b = s[i][j] - '0';	//printf("b = %d\n",b);



				if (s[i][j] <= k && f[b] == 0)

				{

					t++;		//printf("\t t = %d\n",t);

					f[b] = 1;

				}



				if (t == a + 1)

				{

					bleg = 1;

					break;

				}

			}

		}

		if (bleg == 1)

		{

			sum++;

		}

	}

	printf("%d\n", sum);

	return 0;

}