365 A

最新推荐文章于 2021-10-27 04:43:26 发布

japomica

最新推荐文章于 2021-10-27 04:43:26 发布

阅读量773

点赞数

CC 4.0 BY-SA版权

分类专栏： codeforces

本文链接：https://blog.youkuaiyun.com/u010900851/article/details/16924875

codeforces 专栏收录该内容

5 篇文章

订阅专栏

本博客详细解析了一个代码片段，该片段用于找出数组中所有不超过k的k-good数字。通过输入数组长度和k值，代码遍历每个数组元素，判断其是否为k-good数字，并最终输出符合条件的数量。

A. Good Number

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 test(s)

input
10 6
1234560
1234560
1234560
1234560
1234560
1234560
1234560
1234560
1234560
1234560

output
10

input
2 1
1

1
0

output
1

///http://codeforces.com/contest/365/problem/A
#include <iostream>

using namespace std;

int main()
{
    int n,k,a,b;
    while(cin>>n>>k&&n!=0)
    {
        int c=0;//记录个数，切记初始化
        for(int i=0;i<n;i++)
        {
           cin>>a;
           b=0;  
           while(a)
          {
             if(a%10<=k)b|=1<<a%10;   //b=b|(1<<(a%10));位运算
             a/=10;
          }
           if(b+1==2<<k) c++; //2<<k=10000000,如果0...k的值都存在则b=11111111 ,b+1=10000000
        }
        cout<<c<<endl;

    }
}