zoukankan      html  css  js  c++  java
  • HDU6095

    Rikka with Competition

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 772    Accepted Submission(s): 588


    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

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

    Sample Output

    5 1
     

    Source

     
     1 //2017-09-22
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <iostream>
     5 #include <algorithm>
     6 
     7 using namespace std;
     8 
     9 const int N = 110000;
    10 
    11 int n, arr[N], k;
    12 
    13 int main()
    14 {
    15     int T;
    16     scanf("%d", &T);
    17     while(T--){
    18         scanf("%d%d", &n, &k);
    19         for(int i = 0; i < n; i++){
    20             scanf("%d", &arr[i]);
    21         }
    22         sort(arr, arr+n);
    23         int ptr = n-1, ans = 0;
    24         while(ptr >= 1 && arr[ptr]-arr[ptr-1] <= k){
    25             ptr--;
    26             ans++;
    27         }
    28         printf("%d
    ", ans+1);
    29     }
    30 
    31     return 0;
    32 }
  • 相关阅读:
    训练赛(28)—— 计蒜客 45724 Jumping Frog
    训练赛(28)—— 计蒜客 45725 Fujiyama Thursday
    centos上libreoffice+unoconv安装步骤,实现word转pdf
    PhantomJS linux系统下安装步骤及使用方法(网页截屏功能)
    knockout应用开发指南(完整版)
    git 创建版本库
    保留json字符串中文的函数,代替json_encode
    微信公众平台开发接口PHP SDK完整版(转载)
    find_in_set()
    NuSOAP与PHPRPC比较(转)
  • 原文地址:https://www.cnblogs.com/Penn000/p/7580601.html
Copyright © 2011-2022 走看看