HDU 6095 17多校5 Rikka with Competition（思维简单题）

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

 

如果差距在k及以内，两者都有赢的可能。如果差距大于k，大者赢。

因此找出两两差距都在k以内的大数有几个即可。

 

 #include <iostream>
 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<queue>
 #include<cmath>
 #include<string>
 #include<map>
 #include<vector>
 using namespace std;

 int a[];
 bool cmp(int x,int y)
 {
     return x>y;
 }

 int main()
 {
     int T,n,k;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d%d",&n,&k);
         for(int i=;i<n;i++)
             scanf("%d",&a[i]);
         sort(a,a+n,cmp);
         int i;
         for(i=;i<n-;i++)
         {
             if(a[i]-a[i+]>k)
                 break;
         }
         printf("%d\n",i+);
     }
     return ;
 }