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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 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
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;
}
本文标签: 联赛hducompetitionRikka
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