HDU6095 Rikka with Competition(水题，2017 HDU多校联赛 第5场)

题目：

Rikka with Competition

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 246    Accepted Submission(s): 205

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

 

Source

2017 Multi-University Training Contest - Team 5

 

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思路：

先说题意：

给你一组n个数，代表n个选手的能量高低，现在再给你一个k，任意在n个选手中挑取两个选手比赛，如果 |ai−aj|>K那么能量高的选手获胜，另一个将被淘汰，否则两个人都有机会获胜，现在要你求有多少人有获胜的可能

我们只需要排一下序，判断a[i]-a[i+1]>m的个数就可以了

代码：

#include <cstdio>#include <cstring>#include <cctype>#include <string>#include <set>#include <iostream>#include <stack>#include <cmath>#include <queue>#include <vector>#include <algorithm>#define mem(a,b) memset(a,b,sizeof(a))#define inf 0x3f3f3f3f#define N 300#define ll long longusing namespace std;int a[100200];bool cmp(int a,int b){return a>b;}int main(){int t,n,m;scanf("%d",&t);while(t--){int cnt=0;scanf("%d%d",&n,&m);for(int i=0; i<n; i++)scanf("%d",&a[i]);sort(a,a+n,cmp);for(int i=0; i<n; i++){cnt++;if(a[i]-a[i+1]>m)break;}printf("%d\n",cnt);}return 0;}