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
ai without leading zeroes (1 ≤ ai ≤ 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;
}