2017 beijing icpc E - Rikka with Competition
2017-09-22 22:01:19
writer:pprp
A wrestling match will be held tomorrow. nn players will take part in it. The iith player’s strength point is aiai.
If there is a match between the iith player plays and the jjth player, the result will be related to |ai−aj||ai−aj|. If |ai−aj|>K|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−1n−1 matches, the last player will be the winner.
Now, Yuta shows the numbers n,Kn,K and the array aa 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?
InputThe first line contains a number t(1≤t≤100)t(1≤t≤100), the number of the testcases. And there are no more than 22 testcases with n>1000n>1000.
For each testcase, the first line contains two numbers n,K(1≤n≤105,0≤K<109)n,K(1≤n≤105,0≤K<109).
The second line contains nn numbers ai(1≤ai≤109)ai(1≤ai≤109).
OutputFor 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
代码如下:
#include <iostream>
#include <algorithm> using namespace std;
const int maxn = ;
int arr[maxn]; int main()
{
int cas;
cin >> cas;
while(cas--)
{
int n, k;
cin >> n >> k;
int ans = ;
for(int i = ; i < n ; i++)
{
cin >> arr[i];
}
sort(arr,arr+n);
for(int i = n-; i > ; i--)
{
if(arr[i]-arr[i-] > k)
break;
ans++;
}
cout << ans << endl;
} return ;
}
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