2017 beijing icpc E – Rikka with Competition

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