CodeForces - 202C Clear Symmetry （预处理模拟+找规律）

Consider some square matrix A with side n consisting of zeros and ones. There aren rows numbered from 1 to n from top to bottom and n columns numbered from 1 to nfrom 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

4

Output

3

Input

9

Output

5

Note

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

先解释一下这道题的意思啊。。他是说给你一个数，问你在他所给的规则之下，，最少需要几阶方格才能存下。

规则呢：就是说比如一个5阶的情况，刚开始他的所有空都是0。。你要尽可能多的往里面填补1。但是每当你填补一个1以后。

填补位置坐标为i，j。同时n+1-i，j和i，n+1-j也要同时变为1。当另外两个变为1以后。这三个点就可以形成一个正方形，，同理想一下第四个点要变成什么呢。。

没错，也是1。还有就是对n%4之后的结果做处理。。如果结果不为0，在判断是否pos为奇数，否 ，则+1；

如果是奇数时，再看pos -  n  >=  n-1,否，则pos += 2。变成奇数。

还有开始要通过预处理模拟一下各阶能够容纳1的个数。。再看一下x的范围<=100.所以控制其中1的个数大于100就结束就OK啦。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <string>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <map>
#include <sstream>
#include <queue>
#include <stack>
#define INF 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a));
#define For(a,b) for(int i = a;i<b;i++)
#define LL long long
#define MAX_N 100010
using namespace std;
int x[10000];
int ma[1000][1000];
int init()
{
    mem(x,0);
    for(int i = 3; ; i++)
    {
        mem(ma,0);
        int ans = 0;
        for(int j = 1; j<=i; j++)
        {
            for(int k = 1; k<=i; k++)
            {
                if(ma[j][k]) continue;
                if(!ma[j-1][k]&&!ma[j][k-1]&&!ma[j+1][k]&&!ma[j][k+1])
                {
                    ma[j][k] = 1;
                    ans ++;
                    if(!ma[i-j][k]&&!ma[i+1-j][k-1]&&!ma[i+2-j][k]&&!ma[i+1-j][k+1])
                    {
                        if(!ma[i+1-j][k]) ans ++;
                        ma[i+1-j][k] = 1;
                        if(!ma[j-1][i+1-k]&&!ma[j][i-k]&&!ma[j+1][i+1-k]&&!ma[j][i+2-k])
                        {
                            if(!ma[j][i+1-k]) ans ++;
                            ma[j][i+1-k] = 1;
                            if(!ma[i-j][i+1-k]&&!ma[i+1-j][i-k]&&!ma[i+2-j][i+1-k]&&!ma[i+1-j][i+2-k])
                            {
                                if(!ma[i+1-j][i+1-k]) ans ++;
                                ma[i+1-j][i+1-k] = 1;
                            }
                        }
                    }
                }
            }
        }
//        if(i==5)
//            for(int p=1; p<=5; p++)
//            {
//                for(int l = 1; l<=5; l++)cout<<ma[p][l];
//                cout<<endl;
//            }
        x[i] = ans;
        if(ans >= 100) return i;
    }
}
int main()
{
    int num = init();
    int n;
//    for(int i = 3; i<=num; i++)
//    {
//        cout<<i<<' '<<x[i]<<endl;
//    }
    while(~scanf("%d",&n))
    {
        if(n == 1)
        {
            cout<<"1"<<endl;
            continue;
        }
        int pos = lower_bound(x,x+num,n)-x;
        if(n%4)
        {
            n%=4;
            if(pos%2==0) pos ++;
            else if(pos%2 && pos-n < n-1) pos += 2;
        }
        printf("%d\n",pos);
    }
    return 0;
}