HDU 6095 Rikka with Competition(思维)

Rikka with Competition

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 631    Accepted Submission(s): 501

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：给你n个人的power，然后每次抽其中的2个人进行比赛，总共有n-1场比赛，当|ai-aj|>m，power高的人获胜，低的人淘汰，否则两个人都有可能获胜

思路：我们用power最高的人一路打下来，和第二高的人打的时候若|ai-aj|<=m那么第二高的人可能就能赢，我们假定他赢，那他和第三高的打若|ai-aj|<=m，则两个二都有机会赢，那我们假定第三高的赢，那第三高和第四高的打这样一路打下来，那么谁能最后获胜就很清楚了，其实只要样例1能解释清楚，那么这题就清楚了，所以就只要排个序，|ai-aj|<=m的就++当>m就跳出就好了

#include <iostream>#include <stdio.h>#include <algorithm>#include <string.h>using namespace std;bool cmp(int a,int b){    return a>b;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n,k;        scanf("%d%d",&n,&k);        int a[n];        for(int i=0;i<n;i++)            scanf("%d",&a[i]);        sort(a,a+n,cmp);        int sum=0;        for(int i=0;i<n;i++)        {            sum++;            if(a[i]-a[i+1]>k)                break;        }        printf("%d\n",sum);    }    return 0;}