Hdu6095 Rikka with Competition（2017多校第5场）

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本文介绍了一种基于数值差和阈值K的比赛胜者预测算法。该算法通过排序和比较数值差来确定可能获胜的选手数量。适用于n个选手间的淘汰赛制。

Rikka with Competition

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 425 Accepted Submission(s): 350

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

i th player’s strength point is

ai .

If there is a match between the

i th player plays and the

j th 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

Sample Output

5
1

Source

2017 Multi-University Training Contest - Team 5

————————————————————————————————————

题目的意思是给出n个数，进行n-1场比赛，每场比赛如果双方值差的绝对值大于k就大的数赢，否则都有可能赢，问最后最多多少人可能赢

思路：先按大小排序，然后判差值是否大于k，第一次大于k之前的数的个数就是答案

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <unordered_map>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int a[100005];
int T,n,k;
int main()
{

    for(scanf("%d",&T);T--;)
    {
     scanf("%d%d",&n,&k);
     for(int i=0;i<n;i++)
       scanf("%d",&a[i]);
       sort(a,a+n);
       int ans=1;
       for(int i=n-2;i>=0;i--)
       {
           if(a[i+1]-a[i]<=k)
            ans++;
            else
                break;
       }
       printf("%d\n",ans);
    }
    return 0;
}