HDU6095-Rikka with Competition

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Rikka with Competition

                                                               Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                         Total Submission(s): 486    Accepted Submission(s): 398


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 |aiaj|. If |aiaj|>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 n1 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(1t100), 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(1n105,0K<109).

The second line contains n numbers ai(1ai109).
 

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
 

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

题意:先对所有人的能量进行排序,判断出有多少个a[i+1]-a[i]<=k即可



#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 longconst int INF = 0x3f3f3f3f;int a[100005];int n,k;int main(){    int t;    scanf("%d",&t);    while(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;}