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Rikka with Competition

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 396    Accepted Submission(s): 328

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
 
Source 2017 Multi-University Training Contest - Team 5  
Recommend liuyiding   |   We have carefully selected several similar problems for you:   6095  6094  6093  6092  6091   
Statistic |  Submit |  Discuss |  Note 

#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
    int i,j,k,p,t;
    int n;
    int a[100001];
    int tot;
    cin>>t;
    while(t--)
    {
        cin>>n>>k;
        tot=1;
        for(i=1;i<=n;i++)
        {
            cin>>a[i];
        }
        sort(a+1,a+n+1);
        for(i=n-1;i>=1;i--)
        {
            if(a[i+1]-a[i]<=k)tot++;
            else break;
        }
        cout<<tot<<endl;
    }
    return 0;
}


本文标签： 五场competitionhdu6095Rikka

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