【贪心】 Posterized CF980C_resid=1cac1c9786cffed7!980-优快云博客

本文链接：https://blog.youkuaiyun.com/ummmmm/article/details/80291738

本文介绍了一种海报化滤镜算法的实现方法，该算法将图像颜色范围分为连续颜色组，并选择每组的关键颜色来替换像素颜色，以达到保留细节的效果。输入包括图像像素数量及最大组大小，输出是最小字典序的处理后图像。

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http://codeforces.com/problemset/problem/980/C

Professor Ibrahim has prepared the final homework for his algorithm’s class. He asked his students to implement the Posterization Image Filter.

Their algorithm will be tested on an array of integers, where the i-th integer represents the color of the i-th pixel in the image. The image is in black and white, therefore the color of each pixel will be an integer between 0 and 255 (inclusive).

To implement the filter, students are required to divide the black and white color range [0, 255] into groups of consecutive colors, and select one color in each group to be the group’s key. In order to preserve image details, the size of a group must not be greater than k, and each color should belong to exactly one group.

Finally, the students will replace the color of each pixel in the array with that color’s assigned group key.

To better understand the effect, here is an image of a basking turtle where the Posterization Filter was applied with increasing k to the right.

$\text{[math]}$

To make the process of checking the final answer easier, Professor Ibrahim wants students to divide the groups and assign the keys in a way that produces the lexicographically smallest possible array.

Input

The first line of input contains two integers n and k (1≤n≤105, 1≤k≤256), the number of pixels in the image, and the maximum size of a group, respectively.

The second line contains n integers p1,p2,…,pn (0≤pi≤255), where pi is the color of the i-th pixel.

Output

Print n space-separated integers; the lexicographically smallest possible array that represents the image after applying the Posterization filter.

Examples

Input

4 3
2 14 3 4

Output

0 12 3 3

Input

5 2
0 2 1 255 254

Output

0 1 1 254 254

Note

One possible way to group colors and assign keys for the first sample:

Color 2 belongs to the group [0,2], with group key 0.

Color 14 belongs to the group [12,14], with group key 12.

Colors 3 and 4 belong to group [3,5], with group key 3.

Other groups won't affect the result so they are not listed here.

#include <bits/stdc++.h>
using namespace std;

int n, k;
int p[100005];
int id[256];

int main()
{
	while (~scanf ("%d%d", &n, &k) )
	{
		for (int i = 0; i < n; i++)
			scanf ("%d", &p[i]);

		memset (id, -1, sizeof (id) );
		//未分组

		for (int i = 0; i < n; i++)
		{
			if (id[p[i]] == -1) //未分组
			{
				int t = max (0, p[i] - k + 1);  //不可为负
				for (int j = t; j <= p[i]; j++) //可能作为id的值
				{
					if ( (id[j] == -1) || (p[i] - id[j]) < k) //j所在的组可以放进p[i]
					{
						for (int k = j; k <= p[i]; k++)
							id[k] = j;
						break;
					}
				}
			}
		}

		for (int i = 0; i < n; i++)
		{
			printf ("%d", id[p[i]]);
			if (i == n - 1) printf ("\n");
			else printf (" ");
		}
	}
	return 0;
}