E - Rikka with Competition

Posted 2020-10-08 pprp

2017-09-22 22:01:19

writer：pprp

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. nn players will take part in it. The iith player’s strength point is aiai. 

If there is a match between the iith player plays and the jjth player, the result will be related to |ai?aj||ai?aj|. If |ai?aj|>K|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?1n?1 matches, the last player will be the winner. 

Now, Yuta shows the numbers n,Kn,K and the array aa 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?  

InputThe first line contains a number t(1≤t≤100)t(1≤t≤100), the number of the testcases. And there are no more than 22 testcases with n>1000n>1000. 

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

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

OutputFor 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

代码如下：

#include <iostream>
#include <algorithm>

using namespace std;
const int maxn = 110000;
int arr[maxn];

int main()
{
    int cas;
    cin >> cas;
    while(cas--)
    {
        int n, k;
        cin >> n >> k;
        int ans = 1;
        for(int i = 0 ; i < n ; i++)
        {
            cin >> arr[i];
        }
        sort(arr,arr+n);
        for(int i = n-1; i > 0 ; i--)
        {
            if(arr[i]-arr[i-1] > k)
                break;
            ans++;
        }
        cout << ans << endl;
    }

    return 0;
}