hdu 多校联赛 Rikka with Competition

最新推荐文章于 2019-09-21 14:22:06 发布

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最新推荐文章于 2019-09-21 14:22:06 发布

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分类专栏： ACM 杭电多校联赛 ACM题目解析

本文链接：https://blog.youkuaiyun.com/wumingkeqi/article/details/76928385

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本文介绍了一个关于数学竞赛的编程问题，其中涉及n名选手的对决过程及获胜条件，探讨了如何通过编程手段计算可能赢得比赛的选手数量。

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Rikka with Competition

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0

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

Sample Output

5
1

水题交题的时候傻逼了忘了注释sort排序后的测试数据结果错了一次又从头到尾思考了一遍想想都被自己菜哭

ac代码：

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