A. Clear Symmetry
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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 ncolumns 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.
Examples
input
4
output
3
input
9
output
5
Note
The figure below shows the matrices that correspond to the samples:
题意：给你一个数n，表示有n个1，现在你要创造一个n*n的矩阵，把这n个1填到矩阵里，且每两个1之间不能相邻，且构成的矩阵关于横方向和纵方向对称，问所需矩阵最短边长为多少。
心路历程：刚开始找规律被n=3时坑了，3需要特判。
首先由国际棋盘知每个矩阵最多能容下的1的摆法为隔一个摆一个，所以对于奇数矩阵，最多可以摆下（i^2+1）/2, 对于偶数矩阵
最多可以摆下i^2/2，且从试验的结果知，偶数边矩阵无法同时满足对称，不相邻，最少，这三个条件，所以只用讨论奇数边
除了3之外，任何的数都可以由摆下临界值的矩阵摆下。例如，3*3最多摆下5个，5*5可以摆下13个，则5—13的数都可由5*5摆下
ac代码：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int main()
{
int x;
int sum[50];
for(int i=1; i<=50; i+=2) sum[i]=(i*i+1)/2;//先把每个盘所能摆下的1存下来
while(~scanf("%d",&x))
{
if(x==3)//特判3
{
printf("5\n");
continue;
}
for(int i=1; i<=50; i+=2)
{
if(sum[i]>=x)
{
printf("%d\n",i);
break;
}
}
}
return 0;
}