CodeForces 201A Clear Symmetry

 

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 Ai, 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: Ai, j = An - i + 1, j and Ai, j = Ai, 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.

Sample test(s)

input

4

output

3

input

9

outpu5

 

给出了一个矩阵的几个定义(形状，整洁性，均匀性)后，每次给出一个x，求满足形状为x，整洁性，均匀性的n*n的矩阵中，最小的n

 

YY一下，可以知道，n为偶数时一定不是最优解。

当n为奇数时：若x在[l，r]内时n最小都是a[x]行，则在范围[r+1,r+(a[x]+1)*2]内时，n最小都是a[x]+2.

特判1到5，则6~100递推一下就可以了。

 

 

 

 

 

 

 

#include<cstdio>

const int MAXN=105;

int a[MAXN];

int main()
{
    a[1]=1;
    a[3]=5;
    a[2]=a[4]=a[5]=3;
    for(int i=5;i<MAXN;)
    {
        int x=(a[i]+1)*2;
        for(int j=i+1;j<=i+x;j++)
            a[j]=a[i]+2;
        i+=x;
    }

    int x;
    while(scanf("%d",&x)!=EOF)
    {
        printf("%d\n",a[x]);
    }

    return 0;

}

 

Note

 

The figure below shows the matrices that correspond to the samples: