2017 Multi-University Training Contest - Team 5:1011&hdu6095、Rikka with Competition

题目：

[align=left]Problem Description[/align]
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. Aftern−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?  
 

[align=left]Input[/align]
The first line conta
d575
ins a numbert(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).
 

[align=left]Output[/align]
For each testcase, print a single line with a single number -- the answer.
 

[align=left]Sample Input[/align]

2
5 3
1 5 9 6 3
5 2
1 5 9 6 3

 

[align=left]Sample Output[/align]

5
1

题意：n个人比赛，每个人有自己的分值，任意选两个人，如果|a_i-a_j|>k，那么分值大的人获胜，否则两个人都有可能获胜，求有多少人能获胜

思路：签到题。对分值排序。获胜人数处置为1，i从大到小，如果| a[i]-a[i-1] |<=k，获胜人数+1，不然就结束循环，因为继续下去没人能赢当前这个人。

CODE:

#include<bits/stdc++.h>
using namespace std;
int a[100005];
int main()
{
int t,n,k,i;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&k);
for(i=0;i<n;i++) scanf("%d",&a[i]);
sort(a,a+n);
int cnt=1;
for(i=n-1;i>0;i--){
if(a[i]-a[i-1]<=k) cnt++;
else break;
}
printf("%d\n",cnt);
}
return 0;
}