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 test(s)
Input
10 6
1234560
1234560
1234560
1234560
1234560
1234560
1234560
1234560
1234560
1234560
Output
10
Input
2 1
1
10
Output
1
题意:给出n和k,找出n个数中含有0~k这些数字的数有几个
水题,一开始没理解题意,真坑
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std; int main()
{
int n,k,i,ans,j;
char str[1000];
while(~scanf("%d%d",&n,&k))
{
ans = 0;
for(i = 1;i<=n;i++)
{
scanf("%s",str);
int len = strlen(str);
int cnt = 0;
for(j = 0;j<=k;j++)
{
char s[10];
s[0] = j+'0';
s[1] = '\0';
if(!strstr(str,s))
break;
}
if(j>k)
ans++;
}
printf("%d\n",ans);
} return 0;
}