我只是个编程爱好者，有个关于扑克的编程：有1～9 9张扑克，甲抽到的两张之和是10，乙抽到的两张差是1，丙抽到的两张积是24，丁抽到的两张商是3。我自己编了个程序，可就不知道错哪儿了，希望高手指教！高分送上！
/*逻辑推理_扑克 之二*/
#include
void main()
{
int puke[9]={1,2,3,4,5,6,7,8,9};
int zhizhen_a=0,zhizhen_b=1,T=0,erro=0,i,
bijiao_a=0,bijiao_b=0,bijiao_c=0,bijiao_d=0,
he_a=0,he_b=0,cha_a=0,cha_b=0,ji_a=0,ji_b=0,shang_a=0,shang_b,
baocun_hea,baocun_heb,baocun_chaa,baocun_chab,
baocun_jia,baocun_jib;
while(erro==0&&T==0) /*最外层为总循环*/
{
while(erro==0&&T==0&&bijiao_a==0)
{
if(puke[zhizhen_a]+puke[zhizhen_b]==10&&puke[zhizhen_a]&&puke[zhizhen_b]
&&zhizhen_a!=zhizhen_b)
{
he_a=puke[zhizhen_a];
he_b=puke[zhizhen_b];
baocun_hea=zhizhen_a;
baocun_heb=zhizhen_b;
puke[zhizhen_a]=puke[zhizhen_b]=0;
zhizhen_a=0;
while(puke[zhizhen_a]==0)zhizhen_a++;
zhizhen_b=zhizhen_a+1;
bijiao_a=bijiao_c=bijiao_d=1;
bijiao_b=0;
}
else
{
bijiao_a=0;
bijiao_b=bijiao_c=bijiao_d=1;
zhizhen_b++;
if(zhizhen_b>8)
{
zhizhen_a++;
zhizhen_b=zhizhen_a+1;
if(zhizhen_a>8)erro=1;
}
}
}
while(erro==0&&T==0&&bijiao_b==0)
{
if(puke[zhizhen_b]-puke[zhizhen_a]==1&&puke[zhizhen_a]&&puke[zhizhen_b]
&&zhizhen_a!=zhizhen_b)
{
cha_b=puke[zhizhen_b];
cha_a=puke[zhizhen_a];
baocun_chaa=zhizhen_a;
baocun_chab=zhizhen_b;
puke[zhizhen_a]=puke[zhizhen_b]=0;
zhizhen_a=zhizhen_a+1;
while(puke[zhizhen_a]==0)zhizhen_a++;
zhizhen_b=zhizhen_a+1;
bijiao_a=bijiao_b=bijiao_d=1;
bijiao_c=0;
}
else
{
bijiao_b=0;
bijiao_a=bijiao_c=bijiao_d=1;
zhizhen_b++;
if(zhizhen_b>8)
{
zhizhen_a++;
zhizhen_b=zhizhen_a+1;
if(zhizhen_a>8)
{
puke[baocun_hea]=he_a;
puke[baocun_heb]=he_b;
zhizhen_a=baocun_hea+1;
zhizhen_b=zhizhen_a+1;
bijiao_b=bijiao_c=bijiao_d=1;
bijiao_a=0;
}
}
}
}
while(erro==0&&T==0&&bijiao_c==0)
{
if(puke[zhizhen_a]*puke[zhizhen_b]==24&&puke[zhizhen_a]&&puke[zhizhen_b]
&&zhizhen_a!=zhizhen_b)
{
ji_a=puke[zhizhen_a];
ji_b=puke[zhizhen_b];
baocun_jia=zhizhen_a;
baocun_jib=zhizhen_b;
puke[zhizhen_a]=puke[zhizhen_b]=0;
zhizhen_a=0;
while(puke[zhizhen_a]==0)zhizhen_a++;
zhizhen_b=zhizhen_a+1;
bijiao_a=bijiao_b=bijiao_c=1;
bijiao_d=0;
}
else
{
bijiao_c=0;
bijiao_a=bijiao_b=bijiao_d=1;
zhizhen_b++;
if(zhizhen_b>8)
{
zhizhen_a++;
zhizhen_b=zhizhen_a+1;
if(zhizhen_a>8)
{
puke[baocun_chaa]=cha_a;
puke[baocun_chab]=cha_b;
zhizhen_a=baocun_chaa+1;
zhizhen_b=zhizhen_a+1;
bijiao_a=bijiao_c=bijiao_d=1;
bijiao_b=0;
}
}
}
}
while(erro==0&&T==0&&bijiao_d==0)
{
if(puke[zhizhen_a]&&puke[zhizhen_b]&&puke[zhizhen_b]/puke[zhizhen_a]==3
&&puke[zhizhen_b]%puke[zhizhen_a]==0&&zhizhen_a!=zhizhen_b)
{
shang_a=puke[zhizhen_a];
shang_b=puke[zhizhen_b];
puke[zhizhen_a]=puke[zhizhen_b]=0;
T=1;
}
else
{
bijiao_d=0;
bijiao_a=bijiao_b=bijiao_c=1;
zhizhen_b++;
if(zhizhen_b>8)
{
zhizhen_a++;
zhizhen_b=zhizhen_a+1;
if(zhizhen_a>8)
{
puke[baocun_jia]=ji_a;
puke[baocun_jib]=ji_b;
zhizhen_a=baocun_jia+1;
zhizhen_b=zhizhen_a+1;
bijiao_a=bijiao_b=bijiao_d=1;
bijiao_c=0;
}
}
}

}
}
printf("%d %d %d %d %d %d %d %d\n",he_a,he_b,cha_a,cha_b,ji_a,ji_b,shang_a,shang_b);
}

甲:1 9
乙:5 4
丙:3 8
丁:6 2

我也是编程爱好者，不过看你的程序觉头晕啊！
你程序的第一行不对，应该是#include<stdio.h>
但是改过来的程序虽然可以运行，但是结果不对。运用八重循环应该会好一点吧！可以看出结果是
甲:1和9     乙:5和4     丙:3和8    丁:6和2
#include"stdio.h"
main()
{
float p[8];
for(p[0]=1;p[0]<10;p[0]++)
 for(p[1]=1;p[1]<10;p[1]++)
  for(p[2]=1;p[2]<10;p[2]++)
   for(p[3]=1;p[3]<10;p[3]++)
    for(p[4]=1;p[4]<10;p[4]++)
     for(p[5]=1;p[5]<10;p[5]++)
      for(p[6]=1;p[6]<10;p[6]++)
       for(p[7]=1;p[7]<10;p[7]++)
        if(p[0]+p[1]==10&&p[2]-p[3]==1&&p[4]*p[5]==24&&p[6]/p[7]==3&&p[0]!=p[1]&&p[0]!=p[2]&&p[0]!=p[3]&&p[0]!=p[4]&&p[0]!=p[5]
            &&p[0]!=p[6]&&p[0]!=p[7]&&p[1]!=p[2]&&p[1]!=p[3]&&p[1]!=p[4]&&p[1]!=p[5]&&p[1]!=p[6]&&p[1]!=p[7]&&p[2]!=p[3]
            &&p[2]!=p[4]&&p[2]!=p[5]&&p[2]!=p[6]&&p[2]!=p[7]&&p[3]!=p[4]&&p[3]!=p[5]&&p[3]!=p[6]&&p[3]!=p[7]&&p[4]!=p[5]
            &&p[4]!=p[6]&&p[4]!=p[7]&&p[5]!=p[6]&&p[5]!=p[7]&&p[6]!=p[7])
        printf("甲%f和%f，乙%f和%f，丙%f和%f，丁%f和%f\n",p[0],p[1],p[2],p[3],p[4],p[5],p[6],p[7]);
}

来学习下哈。。。

说实话，你给的程序也不是我想要的。我只是想让人帮我找出我程序中的错误，不过你这种思路比我的要简洁的多，所以给你了。谢谢你！

3楼的程序太暴力了

可以用回溯法 不过这颗递归树的规模太大了 c(9,2)*c(8,2)* .... *c(2,2)

既然只有9张牌3楼的方法还是可行的

值得研究！好~