Clear Symmetry CodeForces - 201A

最新推荐文章于 2023-07-08 18:10:34 发布

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最新推荐文章于 2023-07-08 18:10:34 发布

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原文链接：http://www.cnblogs.com/zgglj-com/p/7847049.html

本文探讨了如何寻找最小正整数n，使得存在清晰且对称的n阶矩阵，其元素之和为给定整数x。清晰矩阵是指不包含相邻的1元素，而对称矩阵则指矩阵与其水平或垂直翻转后的形态相同。

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Consider some square matrix A with side n consisting of zeros and ones. There are n rows numbered from 1 to n from top to bottom and n columns numbered from 1 to n from left to right in this matrix. We'll denote the element of the matrix which is located at the intersection of the i-row and the j-th column as A_i, j.

Let's call matrix A clear if no two cells containing ones have a common side.

Let's call matrix A symmetrical if it matches the matrices formed from it by a horizontal and/or a vertical reflection. Formally, for each pair (i, j) (1 ≤ i, j ≤ n) both of the following conditions must be met: A_i, j = A_{n - i + 1, j} and A_i, j = A_{i, n - j + 1}.

Let's define the sharpness of matrix A as the number of ones in it.

Given integer x, your task is to find the smallest positive integer n such that there exists a clear symmetrical matrix A with side n and sharpness x.

Input

The only line contains a single integer x (1 ≤ x ≤ 100) — the required sharpness of the matrix.

Output

Print a single number — the sought value of n.

Example

Input

Output

Input

Output

Note

The figure below shows the matrices that correspond to the samples

题解：①当n是偶数时，n-1所用到1的个数比n多。

　　　②当n是奇数时，最多能用(n*n+1)/2个1。

　　　③能用奇数构成所有的x。

①和③没证明出来，画图归纳的。。。。。

注意：x=1,n=1; x=2,n=3; x=3,n=5;

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 
 4 int x;
 5 int a[]={1,5,13,25,41,61,85,113};
 6 int b[]={1,3,5,7,9,11,13,15};
 7 
 8 int main()
 9 {   cin>>x;
10     int p=0;
11     if(x==3){ cout<<"5"<<endl; return 0; }
12     while(a[p]<x) p++;
13     cout<<b[p]<<endl;
14 }