Codeforces Round #246 (Div. 2)（第一次做cf）

　　早就想做CF，但是在校党，没办法，断网又断电，一天实在忍不住了，就跑着流量做了一次CF

看到A题，看完题就有思路了，然后一把A掉，接着做第二道，看了N久，题意不懂，泪奔啊，然后就没激情再做了，

我的第一次就这样没了。。。。。。。。。。。。。。。。。。。。。。

链接：http://codeforces.com/contest/432/problem/A

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A. Choosing Teams

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The Saratov State University Olympiad Programmers Training Center (SSU OPTC) has n students. For each student you know the number of times he/she has participated in the ACM ICPC world programming championship. According to the ACM ICPC rules, each person can participate in the world championship at most 5 times.

The head of the SSU OPTC is recently gathering teams to participate in the world championship. Each team must consist of exactly three people, at that, any person cannot be a member of two or more teams. What maximum number of teams can the head make if he wants each team to participate in the world championship with the same members at least k times?

Input

The first line contains two integers, n and k (1 ≤ n ≤ 2000; 1 ≤ k ≤ 5). The next line contains n integers: y1, y2, ..., yn (0 ≤ yi ≤ 5), where yi shows the number of times the i-th person participated in the ACM ICPC world championship.

Output

Print a single number — the answer to the problem.

Sample test(s)

input

5 2
0 4 5 1 0

output

1

input

6 4
0 1 2 3 4 5

output

0

input

6 5
0 0 0 0 0 0

output

2

Note

In the first sample only one team could be made: the first, the fourth and the fifth participants.

In the second sample no teams could be created.

In the third sample two teams could be created. Any partition into two teams fits.

代码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;

int main()
{
    int num;
    int n,k;
    int str[2002];
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d",&str[i]);
        }
        sort(str,str+n);
        num=0;
        for(int i=2;i<n;i+=3)
        {
            if((str[i]+k)<=5)
            {
                num++;
            }
        }
        printf("%d
",num);
    }
    return 0;
}