hdu 多校联赛 Rikka with Competition

Rikka with Competition

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

Problem Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:

A wrestling match will be held tomorrow.  n players will take part in it. The  ith player’s strength point is  ai.

If there is a match between the  ith player plays and the  jth player, the result will be related to  |ai−aj|. If  |ai−aj|>K, the player with the higher strength point will win. Otherwise each player will have a chance to win.

The competition rules is a little strange. Each time, the referee will choose two players from all remaining players randomly and hold a match between them. The loser will be be eliminated. After  n−1 matches, the last player will be the winner.

Now, Yuta shows the numbers  n,K and the array  a and he wants to know how many players have a chance to win the competition.

It is too difficult for Rikka. Can you help her?  

 

Input

The first line contains a number  t(1≤t≤100), the number of the testcases. And there are no more than  2 testcases with  n>1000.

For each testcase, the first line contains two numbers  n,K(1≤n≤105,0≤K<109).

The second line contains  n numbers  ai(1≤ai≤109).

 

Output

For each testcase, print a single line with a single number -- the answer.

 

Sample Input

2 5 3 1 5 9 6 3 5 2 1 5 9 6 3

 

Sample Output

5 1

 

水题 交题的时候傻逼了 忘了注释sort排序后的测试数据 结果错了一次 又从头到尾思考了一遍 想想都被自己菜哭

ac代码：

#include
using namespace std;
int main()
{
    int t,n,k;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&k);
        if(n==1)
        {
            printf("1\n");
            continue;
        }
        int a[n];
        memset(a,0,sizeof(a));
        for(int i=0;i0;i--)
        {
            if(a[i]-a[i-1]<=k)
                sum++;
            else break;
        }
        cout<