C语言实现纸牌计算24点小游戏

/*
 *
 * Project : 计算24 
 * 
 * Date : 2015年3月25日10:55:47 
 *
 * Remark ：纸牌中的10以0代替，减法得数均为正数 
 */
 #include <stdio.h>
 #include <stdlib.h>
 #include <math.h>
 #include <time.h>
 
 char getPokerCard(int value)
 {
  if(value==1)
  {
   return 'A';
  }
 else if(value<10)
 {
  return value+'0';
 }
 else if(value==10)
 {
  return '0';
 }
 else if(value==11)
 {
  return 'J';
 }
 else if(value==12)
 {
  return 'Q';
 }
 else if(value==13)
 {
  return 'K';
 }
 }
 
 void printAnswer(int flag,int *poker,char *oper,char *answer)
 {
  char a=getPokerCard(poker[0]);
  char b=getPokerCard(poker[1]);
  char c=getPokerCard(poker[2]);
  char d=getPokerCard(poker[3]);
  
  switch(flag)
  {
   //1.((A*B)*C)*D
   case 1:
    printf("((%c%c%c)%c%c)%c%c\n",a,oper[0],b,oper[1],c,oper[2],d);
    break;
   //2.(A*(B*C))*D
   case 2:
    printf("(%c%c(%c%c%c))%c%c\n",a,oper[0],b,oper[1],c,oper[2],d);
    break;
   //3.(A*B)*(C*D)
   case 3:
    printf("(%c%c%c)%c(%c%c%c)\n",a,oper[0],b,oper[1],c,oper[2],d);
    break;
   //4.A*(B*(C*D))
   case 4:
    printf("%c%c(%c%c(%c%c%c))\n",a,oper[0],b,oper[1],c,oper[2],d);
    break;
   //5.A*((B*C)*D) 
   case 5:
    printf("%c%c((%c%c%c)%c%c)\n",a,oper[0],b,oper[1],c,oper[2],d);
    break;
   default:
    break;
  }
  
  //存储answer
  
  
  return ; 
 }
 
 double getValue(double num1,double num2,char oper)
 {
  double result;
  
  switch(oper)
  {
   case '+':
   result=num1+num2;
   break;
  case '-':
   result=fabs(num1-num2);
   break;
  case '*':
   result=num1*num2;
   break;
  case '/':
   result=num1/num2;
   break;
  default :
   break;
  }
  
  return result;
 }
 
 int getResult(int *poker,char *oper,char *answer)
 {
  double t;
  //将计算值取到 
  int a=poker[0]>10?1:poker[0];
  int b=poker[1]>10?1:poker[1];
  int c=poker[2]>10?1:poker[2];
  int d=poker[3]>10?1:poker[3];
  
  //穷举运算次序
 //1.((A*B)*C)*D
 t=0;
 t=getValue(a,b,oper[0]);
 t=getValue(t,c,oper[1]);
 t=getValue(t,d,oper[2]);
 
 if(fabs(t-24)<0.0001)
 {
  printAnswer(1,poker,oper,answer);
  return 1;
 }
 
 //2.(A*(B*C))*D
 t=0;
 t=getValue(b,c,oper[1]);
 t=getValue(a,t,oper[0]);
 t=getValue(t,d,oper[2]);
 
 if(fabs(t-24)<0.0001)
 {
  printAnswer(2,poker,oper,answer);
  return 1;
 }
 
 //3.(A*B)*(C*D)
 t=0;
 t=getValue(getValue(a,b,oper[0]),getValue(c,d,oper[2]),oper[1]);
 
 if(fabs(t-24)<0.0001)
 {
  printAnswer(3,poker,oper,answer);
  return 1;
 }
 
 //4.A*(B*(C*D))
 t=0;
 t=getValue(c,d,oper[2]);
 t=getValue(b,t,oper[1]);
 t=getValue(a,t,oper[0]);
 
 if(fabs(t-24)<0.0001)
 {
  printAnswer(4,poker,oper,answer);
  return 1;
 }
 
 //5.A*((B*C)*D) 
 t=0;
  t=getValue(b,c,oper[1]);
 t=getValue(t,d,oper[2]);
 t=getValue(a,t,oper[0]);
 
 if(fabs(t-24)<0.0001)
 {
  printAnswer(5,poker,oper,answer);
  return 1;
 }
 
  return 0;
 }
 
 void printResult(int *poker,char *answer)
 {
  char OperKey[4]={'+','-','*','/'};
  char oper[3];
  int i,j,k;
  int count=0;
  
  for(i=0;i<4;i++)
  for(j=0;j<4;j++)
  for(k=0;k<4;k++)
  {
   oper[0]=OperKey[i];
   oper[1]=OperKey[j];
   oper[2]=OperKey[k];
   
   if(getResult(poker,oper,answer))
   count++;
  }
  if(count)
  {
   printf("共%d种解法\n",count);
  }
  else
  {
   printf("该情况无解\n");
  }
  
  return ;
 }
 
 void printPoker(int *poker)
 { 
  int i;
  for(i=0;i<4;i++)
  {
   printf("%c ",getPokerCard(poker[i]));
  }
  printf("\n");
  
  return ;
 }
 
 void getRandomPokers(int *poker)
 {
  int i;
  
  //利用系统时间作为种子产生随机数 ，函数srand()，rand()=>stdlib.h ，函数time()=>time.h 
 srand((unsigned)time(NULL));
 for(i=0;i<4;i++)
 {
  poker[i]=rand()%12+1;
 }
 
 return ; 
 }
 
 main()
 { 
  int poker[4];
 char answer[20];
  char c;
  
  printf("纸牌计算24点\n--------------------------------------------\n");
  do
  {
   //生成随机纸牌 
   getRandomPokers(poker);
   
   printf("随机生成的纸牌为：\n");

   //输出生成的纸牌 
   printPoker(poker);
   
   printf("\n任意键获得结果...\n");
   getchar(); 
   
   //输出计算结果
   printResult(poker,answer);
   
   printf("\n回车键继续，其他键并回车退出...\n");
   c=getchar();
   
   printf("--------------------------------------------\n");
  }
  while(c=='\n');
  
  return 0;
 }