C. The Phone Number

最新推荐文章于 2024-01-18 13:57:19 发布

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最新推荐文章于 2024-01-18 13:57:19 发布

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原文链接：http://www.cnblogs.com/acerkoo/p/9498863.html

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本文探讨了如何构造一个长度为n的排列，使其最长递增子序列(LIS)与最长递减子序列(LDS)长度之和最小。通过将n个元素划分为若干个连续递增的子序列块，并调整块的顺序，实现最小化目标。文章提供了详细的思路解析及代码实现。

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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 ${ai1,ai2,\dots,aikai1,ai2,\dots,aik where 1\leqi1<i2<\dots<ik\leqn1\leqi1<i2<\dots<ik\leqn is called increasing if ai1<ai2<ai3<\dots<aikai1<ai2<ai3<\dots<aik. If ai1>ai2>ai3>\dots>aikai1>ai2>ai3>\dots>aik, a subsequence is called decreasing. An increasing/decreasing subsequence is called longest if it has maximum length among all increasing/decreasing subsequences.}_{{i1,ai2,\dots,aikai1,ai2,\dots,aik where 1\leqi1<i2<\dots<ik\leqn1\leqi1<i2<\dots<ik\leqn is called increasing if ai1<ai2<ai3<\dots<aikai1<ai2<ai3<\dots<aik. If ai1>ai2>ai3>\dots>aikai1>ai2>ai3>\dots>aik, a subsequence is called decreasing. An increasing/decreasing subsequence is called longest if it has maximum length among all increasing/decreasing subsequences.}_{1,ai2,\dots,aikai1,ai2,\dots,aik where 1\leqi1<i2<\dots<ik\leqn1\leqi1<i2<\dots<ik\leqn is called increasing if ai1<ai2<ai3<\dots<aikai1<ai2<ai3<\dots<aik. If ai1>ai2>ai3>\dots>aikai1>ai2>ai3>\dots>aik, a subsequence is called decreasing. An increasing/decreasing subsequence is called longest if it has maximum length among all increasing/decreasing subsequences.}}$

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

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 $nn (1\leqn\leq1051\leqn\leq105) — the length of permutation that you need to build.$

Output

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

If there are multiple answers, print any.

Examples

input

Copy

output

Copy

3 4 1 2

input

Copy

output

Copy

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 $[2,1][2,1]. LIS is [1][1] (or [2][2]), so the length of LIS is equal to 11. LDS is [2,1][2,1], so the length of LDS is equal to 22. The sum is equal to 33. Note that permutation [1,2][1,2] is also valid.$

$题意：给一个数 n，求一个长度为 n 的排列，使得该排列的 lis长度 + lds 长度最小。$

$思路：取 ans = x + y （x 表示 lis 长度，y 表示 lds 长度）那么先将 n 个数字分成 p 块，使每个块中单调递增序列，令 p 块中的数字都为相邻的数。例如 n=9 时，令块一：123、块二：456、块三：789，要是 x + y最小，我们可以得出当排列为块三、块二、块一时最小，即 ans = n/p+p。由数学不等式得： n/p+p>=2*sqrt(n)，即p==sqrt(n)时，ans取最小值 2*sqrt(n) 。由特殊到一般的归纳法，我们即可推出规律。$

$代码如下：$

#include <bits/stdc++.h>
using namespace std;
const int maxn=1e5+10;
int n;
int main(){
    scanf("%d",&n);
    int m=sqrt(n);
    for (int i=0; i<n; i++)
        printf("%d ",max(0,n-m*(i/m+1))+i%m+1);

    return 0;
}