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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 i th player’s strength point is ai .

If there is a match between the i th player plays and the j th 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
  
  
#include<cstdio>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
int a[100010];
int ans;
int main()
{
	int t;
	int n,k;
	scanf("%d",&t);	
	while(t--)
	{
		scanf("%d%d",&n,&k);
		for(int i=0;i<n;i++)
		{
			scanf("%d",&a[i]);
		}
		sort(a,a+n);
		ans=n;
		for(int i=1;i<n;i++)
		{
			if(a[i]-a[i-1]>k)
				ans=n-i;
		}
		printf("%d\n",ans);
	}
	  
}

本文标签： Rikkacompetition

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