Codeforces Round #127 (Div. 1) A. Clear Symmetry 打表

布满荆棘的人生 2022-03-30 05:08 135阅读 0赞

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A. Clear Symmetry

## 题目连接： ##

[http://codeforces.com/contest/201/problem/A][http_codeforces.com_contest_201_problem_A]

## Description ##

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 Input ##

## Sample Output ##

## Hint ##

## 题意 ##

给你一个x，你需要找到一个最小的正方形，使得这个正方形里面有x个1

这个正方形，需要满足1的方格不能相邻，且a\[i\]\[j\]=a\[n-i\]\[j\],a\[i\]\[j\]=a\[i\]\[n-j\]，即上下对称，左右对称

然后问你边长最小是多少

## 题解： ##

数学题 打表

偶数是不考虑的，大概可以画画，很难满足对称性，且中间四个格子是浪费的

然后只用考虑奇数的情况

打表之后发现，奇数我们发现奇数长度的正方形能够容纳的1的个数满足公式2*x*(x+1)+1;

然后套进去就好了

## 代码 ##

#include<bits/stdc++.h>
    using namespace std;
    
    int f(int x)
    {
        x--;
        return 2*x*(x+1)+1;
    }
    int main()
    {
        int n;
        scanf("%d",&n);
        if(n==3)return puts("5");
        int ans = 1;
        while(f(ans)<n)
            ans++;
        cout<<ans*2-1<<endl;
    }

[http_codeforces.com_contest_201_problem_A]: http://codeforces.com/contest/201/problem/A