time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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. 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≤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
Copy
4
output
Copy
3 4 1 2
input
Copy
2
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;
}
原文地址：https://www.cnblogs.com/acerkoo/p/9498863.html