Problem C Team Match


Time Limit: 2000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 160    Accepted Submission(s): 55

Problem Description

The programming competition not only depends on the programmers, but also directed by the coaches. Mr Z is a coach who direct n players to take part in the Guangxi Province Collegiate Programming Contest. We assume that a team is consisted of 3 players whose ability is x, y, z respectively and x >= y >= z. Then the team’s total ability is 3 * x + 2 * y + 1 * z; And for a team, if its ability is not lower than the gold medal level m, the team will certainly win the gold medal. Mr Z would like to match teams to gain as many gold medals as possible, could you tell him how many gold medals it is?


Input

The first line is an integer T which indicates the case number.

And as for each case,  there will be 2 lines.

In the first line, there are 2 integers n m, which indicate the number of players, the gold medal level respectively. Please remember n is always the multiple of 3.

In the second line, there are n integers which represents everyone’s ability.

It is guaranteed that——

T is about 100.

for 100% cases, 1 <= n <= 15, 1 <= m <= 30, 1 <= a[i] <= 20.


Output

As for each case, you need to output a single line.

There should be an integer in the line which means the gold medal teams Mr Z could match.


Sample Input

2

6 18

3 3 3 4 2 2

6 7

1 1 1 1 1 1


Sample Output

2

0


解题思路:从小到大排序后,每次选择一个最小,一个最大,然后中间 a[i](i 从小到大)循环寻找合适的中间数。在还没整个数组统计完时,如果当中有一次 return false,那么就肯定有一对已经没法执行,那么选择最小的尽量靠前的3个淘汰掉,再整个重来。

AC 代码

#include<bits/stdc++.h>
#include<cmath>

#define mem(a,b) memset(a,b,sizeof a);
#define INF 0x3f3f3f3f

using namespace std;

typedef long long ll;
int n,m,a[20];

bool solve(int x)
{
int l=(n/3-x)*3+1, r=n, vis[20];
mem(vis,0);
while(l<r)
{
int flag=0;
for(int i=l+1;i<r;i++)
{
if(!vis[i] && 3*a[r]+2*a[i]+a[l]>=m)
{
flag=1;
vis[i]=1;
break;
}
}

if(!flag) return false;
vis[l]=vis[r]=1;
while(vis[l]) l++;
while(vis[r]) r--;
}
return true;
}

int main()
{
int T; scanf("%d",&T);
while(T-- && ~scanf("%d%d",&n,&m))
{
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
sort(a+1,a+1+n);
int ans=n/3;
while(ans)
{
if(solve(ans)) break;
else ans--;
}
printf("%d\n",ans);
}

return 0;
}