Rikka with Competition（HDU 609）

Rikka with Competition

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 299    Accepted Submission(s): 253

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

25 31 5 9 6 35 21 5 9 6 3

 

Sample Output

51

 

//题意：两个人比赛，如果两人能力值之差大于K，则大的一方获胜，否则两人均有可能获胜，n个人打n-1次比赛，问最后赢的人的可能情况。

//思路：签到题。先给a数组从小到大排个序，sum一开始=n，表示所有人均有可能获胜，然后去遍历a数组，如果相邻两个数的差大于K，那么sum=n-i+1。因为如果相邻有2个数差大于K，那么前面那个人必输，所以能力值比他小的也必输。

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <cmath>#include <algorithm>using namespace std;const int MAX = 1e5 + 100;int n, k;int a[MAX];int main(){int T;scanf("%d", &T);while (T--){scanf("%d%d", &n, &k);for (int i = 1; i <= n; i++)scanf("%d", &a[i]);sort(a + 1, a + n + 1);int sum = n;for (int i = 2; i <= n; i++){if ((a[i] - a[i - 1]) > k){sum = n - i + 1;}}printf("%d\n", sum);}return 0;}