pku 2993 第一周训练——模拟-优快云博客

本文提供了一道来自POJ平台编号为2993的题目解答代码，该题需要构建一个特殊的棋盘，并在指定位置放置棋子。通过字符串解析和坐标映射的方式实现了黑白棋子的放置。

上一周的训练还没做完，唉。只能延续第一周的训练计划了。和pku2996正好过程相反。。。谢模拟题实在是太费劲了。。细节细节注意。

View Code

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
char str[20][40];
int main()
{
int i,j,k,l;
int len;
//首先排除字母构造整个表
    for (i = 0; i < 17; i += 2)
    {
for (j = 0; j < 33; ++j)
        {
if (j%4 == 0)
            str[i][j] = '+';
else
            str[i][j] = '-';
        }
    }
int flag = 1;
for (i = 1; i < 17; i += 2)
    {
if (flag)
        {
            flag = 0;
int pos = 0;
for (j = 0; j <= 28; j += 4)
            {
                str[i][j] = '|';
if (!pos)
                {
                    pos = 1;
for (k = j + 1; k < j + 4; ++k)
                    str[i][k] = '.';
                }
else
                {
                    pos = 0;
for (k = j + 1; k < j + 4; ++k)
                    str[i][k] = ':';
                }
            }
            str[i][j] = '|';
        }
else
        {
            flag = 1;
int pos = 0;
for (j = 0; j <= 28; j += 4)
            {
                str[i][j] = '|';
if (!pos)
                {
                    pos = 1;
for (k = j + 1; k < j + 4; ++k)
                    str[i][k] = ':';
                }
else
                {
                    pos = 0;
for (k = j + 1; k < j + 4; ++k)
                    str[i][k] = '.';
                }
            }
            str[i][j] = '|';
        }
    }
//处理黑白
    char pp[10];
char T[1007];
    scanf("%s",pp);
    scanf("%s",T);
//printf("%s%s\n",pp,T);
    len = strlen(T);
char tmp[4] = {'0'};
    l = 0;
int x,y;
for (i = 0; i < len; ++i)
    {
if (T[i] == ',')
        {
//printf(">>%s\n",tmp);
            if (l == 3)
            {
                y = (tmp[1] - 'a')*4 + 2;
                x = 17 - (tmp[2] - '0')*2;
                str[x][y] = tmp[0];
            }
else if (l == 2)
            {
                y = (tmp[0] - 'a')*4 + 2 ;
                x = 17 - (tmp[1] - '0')*2;
                str[x][y] = 'P';
            }
            l = 0;
            tmp[0] = tmp[1] = tmp[2] = '0';
        }
else
        tmp[l++] = T[i];
    }
if (l == 3)
            {
                 y = (tmp[1] - 'a')*4 + 2;
                x = 17 - (tmp[2] - '0')*2;
                str[x][y] = tmp[0];
            }
else if (l == 2)
            {
                y = (tmp[0] - 'a')*4 + 2;
                x = 17 - (tmp[1] - '0')*2;
                str[x][y] = 'P';
            }
//printf(">%s\n",tmp);
            tmp[0] = tmp[1] = tmp[2] = '0';

    scanf("%s",pp);
    scanf("%s",T);
//printf("%s%s\n",pp,T);
    len = strlen(T);
    l = 0;
for (i = 0; i < len; ++i)
    {
if (T[i] == ',')
        {
//printf("%s\n",tmp);
            if (l == 3)
            {
                y = (tmp[1] - 'a')*4 + 2;
                x = 17 - (tmp[2] - '0')*2;
                str[x][y] = tmp[0] + 32;
            }
else if (l == 2)
            {
                y = (tmp[0] - 'a')*4 + 2;
                x = 17 - (tmp[1] - '0')*2;
                str[x][y] = 'p';
            }
            l = 0;
            tmp[0] = tmp[1] = tmp[2] = '0';
        }
else
        tmp[l++] = T[i];
    }
if (l == 3)
            {
                y = (tmp[1] - 'a')*4 + 2;
                x = 17 - (tmp[2] - '0')*2;
                str[x][y] = tmp[0] + 32;
            }
else if (l == 2)
            {
                y = (tmp[0] - 'a')*4 + 2;
                x = 17 - (tmp[1] - '0')*2;
                str[x][y] = 'p';
            }
//printf("%s\n",tmp);
    tmp[0] = tmp[1] = tmp[2] = '0';
for (i = 0; i < 17; ++i)
    printf("%s\n",str[i]);
return 0;
}