CodeForces - 985E Pencils and Boxes

最新推荐文章于 2020-08-16 16:13:49 发布

转载最新推荐文章于 2020-08-16 16:13:49 发布 · 56 阅读

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原文链接：http://www.cnblogs.com/JYYHH/p/9071227.html

本文介绍了一种将多色铅笔按饱和度差异不超过给定值d且每组至少包含k支铅笔的方式进行分组的算法。通过排序及双指针技巧，判断是否能成功分组。

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可以证明的是，总是存在一种最优策略使得每个组内的权值都是连续的。

所以排完序一遍 two pointers就好啦。

Discription

Mishka received a gift of multicolored pencils for his birthday! Unfortunately he lives in a monochrome world, where everything is of the same color and only saturation differs. This pack can be represented as a sequence a₁, a₂, ..., a_n of ninteger numbers — saturation of the color of each pencil. Now Mishka wants to put all the mess in the pack in order. He has an infinite number of empty boxes to do this. He would like to fill some boxes in such a way that:

Each pencil belongs to exactly one box;
Each non-empty box has at least k pencils in it;
If pencils i and j belong to the same box, then |a_i - a_j| ≤ d, where |x| means absolute value of x. Note that the opposite is optional, there can be pencils iand j such that |a_i - a_j| ≤ d and they belong to different boxes.

Help Mishka to determine if it's possible to distribute all the pencils into boxes. Print "YES" if there exists such a distribution. Otherwise print "NO".

Input

The first line contains three integer numbers n, k and d (1 ≤ k ≤ n ≤ 5·10⁵, 0 ≤ d ≤ 10⁹) — the number of pencils, minimal size of any non-empty box and maximal difference in saturation between any pair of pencils in the same box, respectively.

The second line contains n integer numbers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹) — saturation of color of each pencil.

Output

Print "YES" if it's possible to distribute all the pencils into boxes and satisfy all the conditions. Otherwise print "NO".

Examples

Input

6 3 10
7 2 7 7 4 2

Output

YES

Input

6 2 3
4 5 3 13 4 10

Output

YES

Input

3 2 5
10 16 22

Output

NO

Note

In the first example it is possible to distribute pencils into 2 boxes with 3pencils in each with any distribution. And you also can put all the pencils into the same box, difference of any pair in it won't exceed 10.

In the second example you can split pencils of saturations [4, 5, 3, 4] into 2 boxes of size 2 and put the remaining ones into another box.

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=500005;
int n,a[maxn],K,D,now,L,R;
bool can[maxn];

inline int read(){
	int x=0; char ch=getchar();
	for(;!isdigit(ch);ch=getchar());
	for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
	return x;
}

inline void solve(){
	L=R=0,can[0]=now=1;
	for(int i=1;i<K;i++) can[i]=0;
	for(int i=K;i<=n;i++){
		while(a[i]-a[L+1]>D){ now-=(int)can[L],L++;}
		while(R<i-K){ R++,now+=(int)can[R];}
//		printf("%d %d %d %d\n",i,L,R,now);
		can[i]=now>0?1:0;
	}
}

int main(){
	n=read(),K=read(),D=read();
	for(int i=1;i<=n;i++) a[i]=read();
	sort(a+1,a+n+1);
	
	solve();
	
	puts(can[n]?"YES":"NO");
	return 0;
}

转载于:https://www.cnblogs.com/JYYHH/p/9071227.html