The Phone Number CodeForces - 1017C

最新推荐文章于 2024-01-17 08:41:49 发布

fighting_yifeng

最新推荐文章于 2024-01-17 08:41:49 发布

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本文链接：https://blog.youkuaiyun.com/fighting_yifeng/article/details/81605235

本文介绍了一种通过构造特定排列来解决电话号码价值最小化问题的方法。电话号码的价值定义为其最长递增子序列与最长递减子序列长度之和。文章提供了一个算法，通过将序列分成若干段并适当排序来实现价值最小化。

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Mrs. Smith is trying to contact her husband, John Smith, but she forgot the secret phone number!

The only thing Mrs. Smith remembered was that any permutation of nn can be a secret phone number. Only those permutations that minimize secret value might be the phone of her husband.

The sequence of nn integers is called a permutation if it contains all integers from 11 to nn exactly once.

The secret value of a phone number is defined as the sum of the length of the longest increasing subsequence (LIS) and length of the longest decreasing subsequence (LDS).

A subsequence ai1,ai2,…,aikai1,ai2,…,aik where 1≤i1<i2<…<ik≤n1≤i1<i2<…<ik≤n is called increasing if ai1<ai2<ai3<…<aikai1<ai2<ai3<…<aik . If ai1>ai2>ai3>…>aikai1>ai2>ai3>…>aik , a subsequence is called decreasing. An increasing/decreasing subsequence is called longest if it has maximum length among all increasing/decreasing subsequences.

For example, if there is a permutation [6,4,1,7,2,3,5][6,4,1,7,2,3,5] , LIS of this permutation will be [1,2,3,5][1,2,3,5] , so the length of LIS is equal to 44 . LDS can be [6,4,1][6,4,1] , [6,4,2][6,4,2] , or [6,4,3][6,4,3] , so the length of LDS is 33 .

Note, the lengths of LIS and LDS can be different.

So please help Mrs. Smith to find a permutation that gives a minimum sum of lengths of LIS and LDS.

Input

The only line contains one integer nn (1≤n≤1051≤n≤105 ) — the length of permutation that you need to build.

Output

Print a permutation that gives a minimum sum of lengths of LIS and LDS.

If there are multiple answers, print any.

Examples

Input

Output

3 4 1 2

Input

Output

2 1

Note

In the first sample, you can build a permutation [3,4,1,2][3,4,1,2] . LIS is [3,4][3,4] (or [1,2][1,2] ), so the length of LIS is equal to 22 . LDS can be ony of [3,1][3,1] , [4,2][4,2] , [3,2][3,2] , or [4,1][4,1] . The length of LDS is also equal to 22 . The sum is equal to 44 . Note that [3,4,1,2][3,4,1,2] is not the only permutation that is valid.

In the second sample, you can build a permutation [2,1][2,1] . LIS is [1][1] (or [2][2] ), so the length of LIS is equal to 11 . LDS is [2,1][2,1] , so the length of LDS is equal to 22 . The sum is equal to 33 . Note that permutation [1,2][1,2] is also valid.

分析：

首先是对题意的理解，让我们求一段数字的最长上升和最长下降长度总和。但本题和最长上升子序列应用没半点关系。

具体怎末求呢，我们可以采用分块的方法，将长度为n的序列分为m段。将m段按上升序列排序。则最长上升子序列和最长下降子序列之和。最长上升子序列长度 $\left \lceil n/m \right \rceil$ ，最长下降m。所以总长度m+ $\left \lceil n/m \right \rceil$ 所以我们应分为 $\sqrt{n}$ 块

AC代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string.h>
#include<math.h>
using namespace std;
int main()
{
	int n;
	while(cin >> n)
	{
		int a = sqrt(n);//分a块
		int temp = n, m = n / a;
		for (int i = 0; i < m; i++)
		{
			temp -= a;
			for (int j = temp + 1; j <= temp + a; j++)
				{
					cout << j << " ";
				}
		}
		for (int i = 1; i <= temp; i++)
		{
			cout << i << " ";
		}
		cout << endl;
	}
	return 0;
}