一个C程序设计题,高手请进!

#include
#include

//中间数据，用于生成算式
int g_Order[3][4];//调用顺序
int g_Operator[3];//三次计算的运行符,1-加 2-减 3-倒减 4-乘 5-除 6-倒除
bool Is24(double fNum)
{
double fDif=fNum-24;
if(fDif<0)
fDif=-fDif;

if(fDif<0.00001)
return true;
else
return false;
}

bool Cal2(double fNum1,double fNum2)
{//看2个数是否满足条件
g_Operator[2]=1;
if(Is24(fNum1+fNum2))
return true;
g_Operator[2]=2;
if(Is24(fNum1-fNum2))
return true;
g_Operator[2]=4;
if(Is24(fNum1*fNum2))
return true;
g_Operator[2]=5;
if(fNum2!=0&&Is24(fNum1/fNum2))
return true;
return false;
}

bool Order2(double fNum1,double fNum2)
{
double fTemp[2];
int i,j;
fTemp[0]=fNum1;
fTemp[1]=fNum2;
for(i=0;i<2;i++)
{//数字1从3个中选择一个
for(j=0;j<2;j++)
{//数字2从余下的2个中选一个
if(i!=j)
{
g_Order[2][0]=i;g_Order[2][1]=j;
if(Cal2(fTemp[i],fTemp[j]))
return true;
}
}
}
return false;
}

bool Cal3(double fNum1,double fNum2,double fNum3)
{//看3个数是否满足条件
g_Operator[1]=1;
if(Order2(fNum1+fNum2,fNum3))
return true;
g_Operator[1]=2;
if(Order2(fNum1-fNum2,fNum3))
return true;
g_Operator[1]=4;
if(Order2(fNum1*fNum2,fNum3))
return true;
g_Operator[1]=5;
if(fNum2!=0&&Order2(fNum1/fNum2,fNum3))
return true;
return false;
}

bool Order3(double fNum1,double fNum2,double fNum3)
{
double fTemp[3];
int i,j,k;
fTemp[0]=fNum1;
fTemp[1]=fNum2;
fTemp[2]=fNum3;
for(i=0;i<3;i++)
{//数字1从3个中选择一个
for(j=0;j<3;j++)
{//数字2从余下的2个中选一个
if(i!=j)
{
for(k=0;k<3;k++)
{
if(k!=i&&k!=j)
{
g_Order[1][0]=i;g_Order[1][1]=j;g_Order[1][2]=k;
if(Cal3(fTemp[i],fTemp[j],fTemp[k]))
return true;
}
}
}
}
}
return false;
}

bool Cal4(double fNum1,double fNum2,double fNum3,double fNum4)
{//看4个数是否满足条件
g_Operator[0]=1;
if(Order3(fNum1+fNum2,fNum3,fNum4))
return true;
g_Operator[0]=2;
if(Order3(fNum1-fNum2,fNum3,fNum4))
return true;
g_Operator[0]=4;
if(Order3(fNum1*fNum2,fNum3,fNum4))
return true;
g_Operator[0]=5;
if(fNum2!=0&&Order3(fNum1/fNum2,fNum3,fNum4))
return true;
return false;
}
bool Order4(double fNum1,double fNum2,double fNum3,double fNum4)
{//i,j,k,l四个数字进行组合调用（因为在Cal4中没有考虑到计算的先后，所以在这里面要进行。）
double fTemp[4];
int i,j,k,l;
fTemp[0]=fNum1;
fTemp[1]=fNum2;
fTemp[2]=fNum3;
fTemp[3]=fNum4;
for(i=0;i<4;i++)
{//数字1从4个中选择一个
for(j=0;j<4;j++)
{//数字2从余下的三个中选一个
if(i!=j)
{
for(k=0;k<4;k++)
{//数字3从余下的二个中选一个
if(k!=i&&k!=j)
{
for(l=0;l<4;l++)
{
if(l!=i&&l!=j&&l!=k)
{
g_Order[0][0]=i;g_Order[0][1]=j;g_Order[0][2]=k;g_Order[0][3]=l;
if(Cal4(fTemp[i],fTemp[j],fTemp[k],fTemp[l]))
return true;
}
}
}
}
}
}
}
return false;
}
char GetOper(int nOperator)
{
switch(nOperator)
{
case 1:
return '+';
case 2:
return '-';
case 4:
return '*';
case 5:
return '/';
}
return '\0';
}
void GetCalString(char *cBuf,int i,int j,int k,int l)
{//生成算式
//int g_Order[3][4];//调用顺序
//int g_Operator[3];//三次计算的运行符,1-加 2-减 3-倒减 4-乘 5-除 6-倒除
int nNum[4];
int nTemp[4];
nTemp[0]=i;nTemp[1]=j;nTemp[2]=k;nTemp[3]=l;
nNum[0]=nTemp[g_Order[0][0]];
nNum[1]=nTemp[g_Order[0][1]];
nNum[2]=nTemp[g_Order[0][2]];
nNum[3]=nTemp[g_Order[0][3]];
char cTemp[3][100];
memset(cTemp[0],0,100);
sprintf(cTemp[0],"(%d%c%d)",nNum[0],GetOper(g_Operator[0]),nNum[1]);
nNum[0]=nNum[g_Order[1][0]+1];
nNum[1]=nNum[g_Order[1][1]+1];
nNum[2]=nNum[g_Order[1][2]+1];
memset(cTemp[1],0,100);
if(g_Order[1][0]==0)
sprintf(cTemp[1],"(%s%c%d)",cTemp[0],GetOper(g_Operator[1]),nNum[1]);
else if(g_Order[1][1]==0)
sprintf(cTemp[1],"(%d%c%s)",nNum[0],GetOper(g_Operator[1]),cTemp[0]);
else//g_Order[1][2]==0
sprintf(cTemp[1],"(%d%c%d)",nNum[0],GetOper(g_Operator[1]),nNum[1]);
memset(cTemp[2],0,100);
nNum[0]=nNum[g_Order[2][0]+1];
nNum[1]=nNum[g_Order[2][1]+1];
if(g_Order[1][2]==0)
{//型如:(a+b)-(c+d)
if(g_Order[2][0]==0)
{//没有倒序
sprintf(cTemp[2],"%s%c%s",cTemp[1],GetOper(g_Operator[2]),cTemp[0]);
}
else
{
sprintf(cTemp[2],"%s%c%s",cTemp[0],GetOper(g_Operator[2]),cTemp[1]);
}
}
else
{//其它
if(g_Order[2][0]==0)
{//没有倒序
sprintf(cTemp[2],"%s%c%d",cTemp[1],GetOper(g_Operator[2]),nNum[1]);
}
else
{
sprintf(cTemp[2],"%d%c%s",nNum[0],GetOper(g_Operator[2]),cTemp[1]);
}
}

sprintf(cBuf,"%s=24",cTemp[2]);
}

void main()
{
int i,j,k,l;
int nCount=0,nLossCount=0;
FILE *pFile=::fopen("result.txt","w+");

for(i=1;i<=10;i++)
{
for(j=i;j<=10;j++)
{
for(k=j;k<=10;k++)
{
for(l=k;l<=10;l++)
{
if(Order4(i,j,k,l))
{
//组织运算式
char cTemp[100];//生成运算串
::memset(cTemp,0,100);
GetCalString(cTemp,i,j,k,l);

printf("%2d,%2d,%2d,%2d %s ",i,j,k,l,cTemp);
::fprintf(pFile,"%2d,%2d,%2d,%2d %s ",i,j,k,l,cTemp);
nCount++;
if(nCount%2==0)
{
printf("\r\n");
fprintf(pFile,"\r\n");
}
}
else
nLossCount++;
}
}
}
}

printf("\r\n%d\r\n%d\r\n",nCount,nLossCount);
fprintf(pFile,"\r\n%d\r\n%d\r\n",nCount,nLossCount);
fclose(pFile);
}

我有java实现的24点的程序..可以去我QQ的blog上参考..c的语法和java的差不多吧..只是没有用到你说的那个递归算法..我是全都拿出来的了...
因为我考虑到的是扑克牌没有一个是一样的...就算10 和10也是有黑红梅方的区别吧...

如果不是实际应用的问题还真无聊，
如果是实际应用的问题，劝你绕道，大量浮点可不好玩

去看看http://www.dffy.com/tool/24.htm
看源文件，虽然不是用c写的，但思路很清楚

高手们,是C程序啊.
我不要那些很就以前就贴在网上的java basic程序.能自己劳动一下下帮帮忙吗?
而且那些都不具备我说的那些功能.

留个脚印