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题目:
Rikka with Competition
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 246 Accepted Submission(s): 205
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
2 5 3 1 5 9 6 3 5 2 1 5 9 6 3
Sample Output
5 1
Source 2017 Multi-University Training Contest - Team 5
Recommend liuyiding | We have carefully selected several similar problems for you: 6095 6094 6093 6092 6091
思路:
先说题意:
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
我们只需要排一下序,判断a[i]-a[i+1]>m的个数就可以了
代码:
#include <cstdio>
#include <cstring>
#include <cctype>
#include <string>
#include <set>
#include <iostream>
#include <stack>
#include <cmath>
#include <queue>
#include <vector>
#include <algorithm>
#define mem(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f
#define N 300
#define ll long long
using namespace std;
int a[100200];
bool cmp(int a,int b)
{
return a>b;
}
int main()
{
int t,n,m;
scanf("%d",&t);
while(t--)
{
int cnt=0;
scanf("%d%d",&n,&m);
for(int i=0; i<n; i++)
scanf("%d",&a[i]);
sort(a,a+n,cmp);
for(int i=0; i<n; i++)
{
cnt++;
if(a[i]-a[i+1]>m)
break;
}
printf("%d\n",cnt);
}
return 0;
}
本文标签: 联赛competitionRikkahdu水题
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