hdu 6095
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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 |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
题解:
读题读了好久。贪心题。官方题解。
代码:
#include <bits/stdc++.h>using namespace std;typedef long long LL;const int maxn = 1e5+10;int n;LL k;LL a[maxn];bool cmp(LL c,LL b){ return c>b;}int main(){ int T; cin>>T; while(T--) { cin>>n>>k; for(int i=0;i<n;i++) scanf("%d",&a[i]); sort(a,a+n,cmp); int i; for( i=0;i<n-1;i++) { if(a[i]-a[i+1]>k) {break;} } cout<<i+1<<endl; } return 0;}
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