#include <stdio.h>
#include <math.h>
int count=0;
void empress(int *position)
{
    int i,j,flag;
    while(position[8]!=1)
    {
        ++position[0];
        for(i=0;i<8;++i)
        {
            if(position[i]==8)
            {
                position[i]=0;
                ++position[i+1];
            }
        }
        flag=1;//判断结果是否满足条件
        for(i=0;i<8;++i)
        {
            for(j=0;j<8;++j)
            {
                if(i!=j)
                {
                    if(position[i]==position[j])
                        flag=0;
                    else if(abs(position[i]-position[j])==abs(i-j))
                        flag=0;
                }
            }
        }
        if(flag==1)//输出#代表皇后                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                            
        {
            for(i=0;i<8;++i)
            {
                for(j=0;j<8;++j)
                {
                    if(position[i]==j)
                        printf("1");
                    else
                        printf("0");
                }
                printf("\n");
            }
            printf("\n");
        }
        count+=flag;
    }
}
int main()
{
    int position[9]={0};
    empress(position);
    printf("%d种解\n",count);
    return 0;
}

#include <stdio.h>
#include <math.h>

//定义全局变量，用以计数
 int count=0;

 //主算法（递归思想）
 void empress(int *position)
 {
     //i,j是循环控制变量，flag是开关变量
     int i,j,flag;

     //这段我再看看
     while(position[8]!=1)
     {
         ++position[0];
         for(i=0;i<8;++i)
         {
             if(position[i]==8)
             {
                 position[i]=0;
                 ++position[i+1];
             }
         }
         flag=1;

         //判断是否形成攻击
        for(i=0;i<8;++i)
         {
             for(j=0;j<8;++j)
             {
                 //两皇后在不同行的时候
                 if(i!=j)
                 {
                     //如果在同一列
                     if(position[i]==position[j])
                         flag=0;//形成了攻击，改变开关变量（off）

                     //如果在同一对角线
                     else if(abs(position[i]-position[j])==abs(i-j))
                         flag=0;//形成攻击，改变开关变量（off）
                 }
             }
         }

        
         if(flag==1)//输出1代表皇后                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                            
         {
             //对结果进行打印，就是把一维的数据对应到一个二维的表现形式上，这部分不难。
             for(i=0;i<8;++i)
             {
                 for(j=0;j<8;++j)
                 {
                     if(position[i]==j)
                         printf("1");
                     else
                         printf("0");
                 }
                 printf("\n");
             }
             printf("\n");
         }
         count+=flag;
     }
 }
 int main()
 {
     //定义一个一维数组用来存放每一列上皇后在第几行
     //首先你要明确一点，在8*8的布局上，每一列是有且仅有一个皇后的
     //所以你只需要记录每一列皇后在第几行上就ok了。
     //所以用一个一维的数组解决这个问题。
     int position[9]={0};

     empress(position);

     printf("%d种解\n",count);

     return 0;
 }

竟然忘了给你继续解释了。。

罪过罪过。。

https://bbs.bccn.net/thread-414535-1-1.html

里面找找，有八皇后的代码。