C++博客-familyhttp://www.cppblog.com/family/zh-cnSat, 04 Apr 2026 05:41:44 GMTSat, 04 Apr 2026 05:41:44 GMT60关于五子棋代码。求指导！！！http://www.cppblog.com/family/archive/2012/04/12/family.html迎风而立迎风而立Thu, 12 Apr 2012 15:48:00 GMThttp://www.cppblog.com/family/archive/2012/04/12/family.htmlhttp://www.cppblog.com/family/comments/171162.htmlhttp://www.cppblog.com/family/archive/2012/04/12/family.html#Feedback5http://www.cppblog.com/family/comments/commentRss/171162.htmlhttp://www.cppblog.com/family/services/trackbacks/171162.html人人对战：


#include<iostream>
using namespace std;
void wz()
{
 cout<<"下面开始五子棋比赛，一号选手用方块符号，二号用心型符号:"<<endl<<endl;
 for(int i=3;i>0;i--)
 {
  cout<<i<<"  ";
 }
 cout<<"开始-----------"<<endl<<endl;
}

int hs()
{
 int x,y,i=0,j=1;
       char d[80][80];
    //初始化---
       for(x=0;x<25;x+=3)
       {
              for(y=0;y<80;y+=3)
              {
                     d[x][y]='+';
     }
    }

     //第一行数字--
    for(x=0;x<10;x++)
           cout<<x<<"  ";
    for(x=10;x<27;x++)
     cout<<x<<" ";
    cout<<endl;

//输出棋盘--
    for(x=0;x<25;x+=3)
       {
     cout<<j<<"  ";j++;
              for(y=0;y<76;y+=3)
              {
     
                     cout<<d[x][y]<<"  ";
     }
     cout<<endl;
    }
   

 
  

 while(1)
 {
  
  int a,b;
  if(i%2==0)
  cout<<"由玩家一下棋："<<endl;
  else
   cout<<"由玩家二下棋："<<endl;

  cin>>a>>b;
  if(i%2==1)
  {
   if(d[(a-1)*3][(b-1)*3]=='+')
         d[(a-1)*3][(b-1)*3]=3;
   else
   {
    cout<<"-----------------------------不能覆盖棋子:"<<endl;
    continue;
   }
  }
  else
  {
   if(d[(a-1)*3][(b-1)*3]=='+')
   d[(a-1)*3][(b-1)*3]=1;
   else
   {
    cout<<"-----------------------------不能覆盖棋子:"<<endl;
    continue;
   }
  }
  i++;
  j=1;
  //第一行数字---
 for(x=0;x<10;x++)
           cout<<x<<"  ";
    for(x=10;x<27;x++)
     cout<<x<<" ";
    cout<<endl;

    for(x=0;x<25;x+=3)
       {
     cout<<j<<"  ";j++;
              for(y=0;y<76;y+=3)
              {
     
                     cout<<d[x][y]<<"  ";
     }
     cout<<endl;
    }
//判断是否是五个
    if(//横着
     d[(a-1)*3][(b-1)*3]==1&&d[(a-1)*3][(b)*3]==1&&d[(a-1)*3][(b+1)*3]==1&&d[(a-1)*3][(b+2)*3]==1&&d[(a-1)*3][(b+3)*3]==1
     ||d[(a-1)*3][(b-2)*3]==1&&d[(a-1)*3][(b-1)*3]==1&&d[(a-1)*3][(b)*3]==1&&d[(a-1)*3][(b+1)*3]==1&&d[(a-1)*3][(b+2)*3]==1
     ||d[(a-1)*3][(b-3)*3]==1&&d[(a-1)*3][(b-2)*3]==1&&d[(a-1)*3][(b-1)*3]==1&&d[(a-1)*3][(b)*3]==1&&d[(a-1)*3][(b+1)*3]==1
     ||d[(a-1)*3][(b-4)*3]==1&&d[(a-1)*3][(b-3)*3]==1&&d[(a-1)*3][(b-2)*3]==1&&d[(a-1)*3][(b-1)*3]==1&&d[(a-1)*3][(b)*3]==1
     ||d[(a-1)*3][(b-5)*3]==1&&d[(a-1)*3][(b-4)*3]==1&&d[(a-1)*3][(b-3)*3]==1&&d[(a-1)*3][(b-2)*3]==1&&d[(a-1)*3][(b-1)*3]==1
     //从左上到右下
     ||d[(a-1)*3][(b-1)*3]==1&&d[a*3][b*3]==1&&d[(a+1)*3][(b+1)*3]==1&&d[(a+2)*3][(b+2)*3]==1&&d[(a+3)*3][(b+3)*3]==1
     ||d[(a-2)*3][(b-2)*3]==1&&d[(a-1)*3][(b-1)*3]==1&&d[a*3][b*3]==1&&d[(a+1)*3][(b+1)*3]==1&&d[(a+2)*3][(b+2)*3]==1
     ||d[(a-3)*3][(b-3)*3]==1&&d[(a-2)*3][(b-2)*3]==1&&d[(a-1)*3][(b-1)*3]==1&&d[a*3][b*3]==1&&d[(a+1)*3][(b+1)*3]==1
     ||d[(a-4)*3][(b-4)*3]==1&&d[(a-3)*3][(b-3)*3]==1&&d[(a-2)*3][(b-2)*3]==1&&d[(a-1)*3][(b-1)*3]==1&&d[a*3][b*3]==1
     ||d[(a-5)*3][(b-5)*3]==1&&d[(a-4)*3][(b-4)*3]==1&&d[(a-3)*3][(b-3)*3]==1&&d[(a-2)*3][(b-2)*3]==1&&d[(a-1)*3][(b-1)*3]==1
     //从右上到左下
     ||d[(a-1)*3][(b-1)*3]==1&&d[(a-2)*3][b*3]==1&&d[a-3][b+1]==1&&d[a-4][b+2]==1&&d[a-5][b+3]==1
     ||d[a*3][(b-2)*3]==1&&d[(a-1)*3][(b-1)*3]==1&&d[(a-2)*3][b*3]==1&&d[(a-3)*3][(b+1)*3]==1&&d[(a-4)*3][(b+2)*3]==1
     ||d[(a+1)*3][(b-3)*3]==1&&d[a*3][(b-2)*3]==1&&d[(a-1)*3][(b-1)*3]==1&&d[(a-2)*3][b*3]==1&&d[(a-3)*3][(b+1)*3]==1
     ||d[(a+2)*3][(b-4)*3]==1&&d[(a+1)*3][(b-3)*3]==1&&d[a*3][(b-2)*3]==1&&d[(a-1)*3][(b-1)*3]==1&&d[(a-2)*3][b*3]==1
     ||d[(a+3)*3][(b-5)*3]==1&&d[(a+2)*3][(b-4)*3]==1&&d[(a+1)*3][(b-3)*3]==1&&d[a*3][(b-2)*3]==1&&d[(a-1)*3][(b-1)*3]==1
     //竖着
        ||d[(a-1)*3][(b-1)*3]==1&&d[a*3][(b-1)*3]==1&&d[(a+1)*3][(b-1)*3]==1&&d[(a+2)*3][(b-1)*3]==1&&d[(a+3)*3][(b-1)*3]==1
     ||d[(a-2)*3][(b-1)*3]==1&&d[(a-1)*3][(b-1)*3]==1&&d[a*3][(b-1)*3]==1&&d[(a+1)*3][(b-1)*3]==1&&d[(a+2)*3][(b-1)*3]==1
     ||d[(a-3)*3][(b-1)*3]==1&&d[(a-2)*3][(b-1)*3]==1&&d[(a-1)*3][(b-1)*3]==1&&d[a*3][(b-1)*3]==1&&d[(a+1)*3][(b-1)*3]==1
     ||d[(a-4)*3][(b-1)*3]==1&&d[(a-3)*3][(b-1)*3]==1&&d[(a-2)*3][(b-1)*3]==1&&d[(a-1)*3][(b-1)*3]==1&&d[a*3][(b-1)*3]==1
     ||d[(a-5)*3][(b-1)*3]==1&&d[(a-4)*3][(b-1)*3]==1&&d[(a-3)*3][(b-1)*3]==1&&d[(a-2)*3][(b-1)*3]==1&&d[(a-1)*3][(b-1)*3]==1
     )
    {
     cout<<endl<<"玩家一获胜"<<endl;break;
    }

    if(//横着
     d[(a-1)*3][(b-1)*3]==3&&d[(a-1)*3][(b)*3]==3&&d[(a-1)*3][(b+1)*3]==3&&d[(a-1)*3][(b+2)*3]==3&&d[(a-1)*3][(b+3)*3]==3
     ||d[(a-1)*3][(b-2)*3]==3&&d[(a-1)*3][(b-1)*3]==3&&d[(a-1)*3][(b)*3]==3&&d[(a-1)*3][(b+1)*3]==3&&d[(a-1)*3][(b+2)*3]==3
     ||d[(a-1)*3][(b-3)*3]==3&&d[(a-1)*3][(b-2)*3]==3&&d[(a-1)*3][(b-1)*3]==3&&d[(a-1)*3][(b)*3]==3&&d[(a-1)*3][(b+1)*3]==3
     ||d[(a-1)*3][(b-4)*3]==3&&d[(a-1)*3][(b-3)*3]==3&&d[(a-1)*3][(b-2)*3]==3&&d[(a-1)*3][(b-1)*3]==3&&d[(a-1)*3][(b)*3]==3
     ||d[(a-1)*3][(b-5)*3]==3&&d[(a-1)*3][(b-4)*3]==3&&d[(a-1)*3][(b-3)*3]==3&&d[(a-1)*3][(b-2)*3]==3&&d[(a-1)*3][(b-1)*3]==3
     //从左上到右下
     ||d[(a-1)*3][(b-1)*3]==3&&d[a*3][b*3]==3&&d[(a+1)*3][(b+1)*3]==3&&d[(a+2)*3][(b+2)*3]==3&&d[(a+3)*3][(b+3)*3]==3
     ||d[(a-2)*3][(b-2)*3]==3&&d[(a-1)*3][(b-1)*3]==3&&d[a*3][b*3]==3&&d[(a+1)*3][(b+1)*3]==3&&d[(a+2)*3][(b+2)*3]==3
     ||d[(a-3)*3][(b-3)*3]==3&&d[(a-2)*3][(b-2)*3]==3&&d[(a-1)*3][(b-1)*3]==3&&d[a*3][b*3]==3&&d[(a+1)*3][(b+1)*3]==3
     ||d[(a-4)*3][(b-4)*3]==3&&d[(a-3)*3][(b-3)*3]==3&&d[(a-2)*3][(b-2)*3]==3&&d[(a-1)*3][(b-1)*3]==3&&d[a*3][b*3]==3
     ||d[(a-5)*3][(b-5)*3]==3&&d[(a-4)*3][(b-4)*3]==3&&d[(a-3)*3][(b-3)*3]==3&&d[(a-2)*3][(b-2)*3]==3&&d[(a-1)*3][(b-1)*3]==3
     //从右上到左下
     ||d[(a-1)*3][(b-1)*3]==3&&d[(a-2)*3][b*3]==3&&d[a-3][b+1]==3&&d[a-4][b+2]==3&&d[a-5][b+3]==3
     ||d[a*3][(b-2)*3]==3&&d[(a-1)*3][(b-1)*3]==3&&d[(a-2)*3][b*3]==3&&d[(a-3)*3][(b+1)*3]==3&&d[(a-4)*3][(b+2)*3]==3
     ||d[(a+1)*3][(b-3)*3]==3&&d[a*3][(b-2)*3]==3&&d[(a-1)*3][(b-1)*3]==3&&d[(a-2)*3][b*3]==3&&d[(a-3)*3][(b+1)*3]==3
     ||d[(a+2)*3][(b-4)*3]==3&&d[(a+1)*3][(b-3)*3]==3&&d[a*3][(b-2)*3]==3&&d[(a-1)*3][(b-1)*3]==3&&d[(a-2)*3][b*3]==3
     ||d[(a+3)*3][(b-5)*3]==3&&d[(a+2)*3][(b-4)*3]==3&&d[(a+1)*3][(b-3)*3]==3&&d[a*3][(b-2)*3]==3&&d[(a-1)*3][(b-1)*3]==3
     //竖着
        ||d[(a-1)*3][(b-1)*3]==3&&d[a*3][(b-1)*3]==3&&d[(a+1)*3][(b-1)*3]==3&&d[(a+2)*3][(b-1)*3]==3&&d[(a+3)*3][(b-1)*3]==3
     ||d[(a-2)*3][(b-1)*3]==3&&d[(a-1)*3][(b-1)*3]==3&&d[a*3][(b-1)*3]==3&&d[(a+1)*3][(b-1)*3]==3&&d[(a+2)*3][(b-1)*3]==3
     ||d[(a-3)*3][(b-1)*3]==3&&d[(a-2)*3][(b-1)*3]==3&&d[(a-1)*3][(b-1)*3]==3&&d[a*3][(b-1)*3]==3&&d[(a+1)*3][(b-1)*3]==3
     ||d[(a-4)*3][(b-1)*3]==3&&d[(a-3)*3][(b-1)*3]==3&&d[(a-2)*3][(b-1)*3]==3&&d[(a-1)*3][(b-1)*3]==3&&d[a*3][(b-1)*3]==3
     ||d[(a-5)*3][(b-1)*3]==3&&d[(a-4)*3][(b-1)*3]==3&&d[(a-3)*3][(b-1)*3]==3&&d[(a-2)*3][(b-1)*3]==3&&d[(a-1)*3][(b-1)*3]==3
     )
    {
   cout<<endl<<"玩家二获胜"<<endl;break;
    }
 }
 return 0;
}
int main()
{
 char a;
 wz();
 hs();
 while(1)
  {
  cout<<"按P重新开始：";
  cin>>a;
  if(a=='p')
   hs();
  else
   break;
  }
 return 0; 
}




人机对战暂时没想出来，有知道的希望指教一下！！感激----



迎风而立 2012-04-12 23:48 发表评论
]]>关于移位运算符和取反运算符的一点感悟http://www.cppblog.com/family/archive/2012/04/08/famliy.html迎风而立迎风而立Sun, 08 Apr 2012 13:29:00 GMThttp://www.cppblog.com/family/archive/2012/04/08/famliy.htmlhttp://www.cppblog.com/family/comments/170472.htmlhttp://www.cppblog.com/family/archive/2012/04/08/famliy.html#Feedback2http://www.cppblog.com/family/comments/commentRss/170472.htmlhttp://www.cppblog.com/family/services/trackbacks/170472.html今天在学习移位运算符和取反运算符的时候，发现一些问题！现在呢在这里说一下自己的感悟！

先看下面一道 取反和移位相结合的问题


#include<stdio.h>
int main()
{
 char a=0x11;
 char b;
 b=~a<<2;
 printf("%d\n",b);
 b=~(a<<2);
 printf("%d\n",b);
 b=~a>>2;
 printf("%d\n",b);
 b=~(a>>2);
 printf("%d\n",b);
 return 0;
}

 

输出结果是：

-72
-69
-5
-5

 

刚开始在想为什么相同的操作（只是移位方向不同），为什么上面两个结果不一样，而下面两个结果却是一样的呢。有些人在疑问是不是因为下面两个表达式都是先移位后取反的呢？我可以肯定不是这样的，但是不能解释这个结果。这个问题我纠结了半天！  不过对下面的知识了解之后便豁然开朗了。

移位运算符分为左移运算符和右移运算符。
当对一个数执行左移运算的时候，右边的空位是用0补齐。
而当对一个数执行右移运算的时候，左边的空位是用符号位补齐的（这是我之前没注意的地方，惭愧）。

 

解体步骤如下：


a=0x11=0001 0001

~a=1110 1110
a<<2=0100 0100
a>>2=0000 0100


~a<<2=1011 1000 
现在 1011 1000是补码，转化成原码之后是
1100 1000=-72  -72是十进制

~（a<<2）=0100 0100
转化为原码之后是 1100 0101=-69


~a>>2=1111 1011(注意现在是用1补齐的，因为~a的符号位是1)
转化为原码之后是  1000 0101=-5


~（a>>2)=1111 1011
转化为原码之后是  1000 0101=-5


明白之后心情很好，所以在这里分享一下。呵呵



迎风而立 2012-04-08 21:29 发表评论
]]>