codeforces 384A Coder

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本文探讨了如何在n×n的棋盘上放置名为Coder的新棋子，使其不互相攻击的问题。Coder可以水平或垂直移动一格进行攻击。文章提供了一个C++实现方案，通过模拟的方式找出最大数量的Coder放置配置。

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Iahub likes chess very much. He even invented a new chess piece named Coder. A Coder can move (and attack) one square horizontally or vertically. More precisely, if the Coder is located at position (x, y), he can move to (or attack) positions (x + 1, y), (x–1, y), (x, y + 1)and (x, y–1).

Iahub wants to know how many Coders can be placed on an n × n chessboard, so that no Coder attacks any other Coder.

Input

The first line contains an integer n (1 ≤ n ≤ 1000).

Output

On the first line print an integer, the maximum number of Coders that can be placed on the chessboard.

On each of the next n lines print n characters, describing the configuration of the Coders. For an empty cell print an '.', and for a Coder print a 'C'.

If there are multiple correct answers, you can print any.

        Examples 
      
          input 
        
2

          output 
        
2
C.
.C

模拟

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    scanf("%d", &n);
    char s[1100][1100];
    memset(s, '.', sizeof(s));
    int sum = 0;
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= n; j++)
        {
            bool flag = true;
            int p = i - 1, q = i + 1;
            if(p >= 1 && p <= n && s[p][j] == 'C')
            {
                flag = false;
            }
            if(q >= 1 && q <= n && s[q][j] == 'C')
            {
                flag = false;
            }
            p = j - 1;
            q = j + 1;
            if(p >= 1 && p <= n && s[i][p] == 'C')
            {
                flag = false;
            }
            if(p >= 1 && p <= n && s[i][q] == 'C')
            {
                flag = false;
            }
            if(flag)
            {
                s[i][j] = 'C';
                sum++;
            }
        }
    }
    printf("%d\n", sum);
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= n; j++)
        {
            printf("%c", s[i][j]);
        }
        printf("\n");
    }
    return 0;
}