CF 246 A Choosing Teams

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本文探讨了一种基于学生参与ACM国际大学生程序设计竞赛次数的团队选拔算法。该算法旨在最大化组建能在至少k次比赛中保持不变的三人民间队伍数量。通过对输入数据排序并检查每三个一组的学生参赛资格，可以高效地计算出可能的最大队伍数。

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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

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <ctype.h>
#include <iostream>
#include <algorithm>
#include <map>
#include <string>
#include <vector>

using namespace std;

#define Cmp(a,b) strcmp(a,b)
#define Copy(a,b) strcpy(a,b);
#define Len(a) strlen(a);
#define sn getchar()
#define fr(a) for(int i=0;i<n;i++)
#define fn(a) for(int i=1;i<=n;i++)
#define f1(a) for(int i=1;i<n;i++)
#define pn cout<<endl
#define pc(a) printf("Case %d:",a)
#define pr cout<<" "
#define MEM(a) memset(a,0,sizeof(a))
#define MEMF(a) memset(a,false,sizeof(a))
#define w1 while(1)
#define w(a) while(a--)
#define INF 4294967295
#define PI 3.14159265359

const int dir4[4][2]= {{-1,0},{1,0},{0,-1},{0,1}};
const int dir8[8][2]= {{-1,0},{1,0},{0,-1},{0,1},{-1,1},{1,-1},{-1,-1},{1,1}};

/***********************************************/

int main()
{
    int n,k;
    int peo[2001];
    MEM(peo);
    cin>>n>>k;
    for(int i=0;i<n;i++)
    {
        cin>>peo[i];
    }
    int sum=0;
    sort(peo,peo+n);
    for(int i=2;i<n;i+=3)
    {
        if(peo[i]+k<=5)
            sum++;
    }
    cout<<sum<<endl;;
    return 0;
}