E. Pencils and Boxes（树状数组+dp）

E. Pencils and Boxes

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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, ..., an of n integer 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 |ai - aj| ≤ d, where |x| means absolute value of x. Note that the opposite is optional, there can be pencils i and j such that |ai - aj| ≤ 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, ..., an (1 ≤ ai ≤ 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

Copy

6 3 10
7 2 7 7 4 2

output

Copy

YES

input

Copy

6 2 3
4 5 3 13 4 10

output

Copy

YES

input

Copy

3 2 5
10 16 22

output

Copy

NO

题目大概：

给出n个数，要求给这些数分组，每组不少于k个。每组的数之间的差不能大于d。是否能够分组？

思路：

可以考虑排序后，用dp。dp【i】代表1到i是否可以分组成功。

dp【i】能否分组，取决于dp【j】到dp【i-k+1】；dp【j】代表最小的一个与i差值小于d的。

如果dp【j】到dp【i-k+1】有一个是可以分组成功的，那dp【i】也可以分组成功。

但是如果直接两重for求解的话，会超时。所以，用树状数组来统计一下1到n可以分组的数的数量。

代码：

#include 

using namespace std;
const int maxn=500005;
int a[maxn];
int dp[maxn];
int c[maxn];
int lowbit(int x)
{
    return x&(-x);
}
void add(int x,int v)
{
    while(x<=maxn)
    {
        c[x]+=v;
        x+=lowbit(x);
    }

}
int sum(int x)
{
    int sum=0;
    while(x>0)
    {
        sum+=c[x];
        x-=lowbit(x);
    }
    return sum;
}
int get(int l,int r)
{
    if(l>r)return 0;
    else return sum(r)-sum(l-1);
}
int main()
{
    int n,k,d;
    scanf("%d%d%d",&n,&k,&d);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
    }
    sort(a+1,a+n+1);
    dp[1]=1;
    add(1,1);
    int l=1;
    for(int i=2;i<=n;i++)
    {
        while(ld)++l;
        dp[i+1]=(get(l,i-k+1)>=1);
        if(dp[i+1])add(i+1,dp[i+1]);
    }
    if(dp[n+1])
    {
        printf("YES\n");
    }
    else printf("NO\n");
    return 0;
}