CodeForces 384A Coder

最新推荐文章于 2020-09-25 15:43:19 发布

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最新推荐文章于 2020-09-25 15:43:19 发布

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原文链接：http://www.cnblogs.com/cyd308/p/4771513.html

解决在棋盘上放置码器的问题，使得没有码器可以攻击到另一个码器，包括输入输出规范和代码实现。

Coder

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 384A

Description

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.

Sample Input

Input

Output

2
C.
.C

 1 #include <stdio.h>
 2 #include <string.h>
 3 int main()
 4 {
 5     int n;
 6     int i,j,k;
 7     while(scanf("%d",&n)!=EOF)
 8     {
 9         if(n%2==0)
10         {
11             printf("%d\n",n*n/2);
12             for(i=1;i<=n/2;i++)
13             {
14                 for(j=1;j<=n;j++)
15                     if(j%2==1)
16                         printf("C");
17                     else
18                         printf(".");
19                 printf("\n");
20                 for(j=1;j<=n;j++)
21                     if(j%2==0)
22                         printf("C");
23                     else
24                         printf(".");
25                 printf("\n");
26             }
27         }
28         else
29         {
30             printf("%d\n",(n+1)*(n+1)/4+(n-1)*(n-1)/4);
31             for(i=1;i<=n;i++)
32             {
33                 if(i%2==1)
34                 {
35                     for(j=1;j<=n;j++)
36                         if(j%2==1)
37                             printf("C");
38                         else
39                             printf(".");
40                     printf("\n");
41                 }
42                 else
43                 {
44                     for(j=1;j<=n;j++)
45                         if(j%2==0)
46                             printf("C");
47                         else
48                             printf(".");
49                     printf("\n");
50                 }
51             }
52         }
53     }
54     return 0;
55 }