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主题:做了个算24点牌的C程序,另求算“猜数字”思想

先说“猜数字”,是不是最少7步?

//以下为24点程序//
//    +    -    *    /    --    -/
//    0    1    2    3    4    5

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

int treat(float a,float b,float c,float d);
float myF(int flag,float m,float n);
void myPrint(int type,int i,int j,int k,float a,float b,float c,float d);

int time,temp=0;
void main()
{
    int i,j,k,t,again,res,flag;
    float num[4];
    again=1;
    while(again==1)
    {
        printf ("\nPlease Enter 4 nums(1~13):\n");
        i=0;
        flag=0;
        while (flag==0)
        {
            i++;
        //    printf ("Input num-%d\n",i);
            for(i=0;i<4;i++)
            {
                scanf("%f",&num[i]);
                if (num[i]<1 || num[i]>13 || num[i]!=int(num[i]))
                    flag++;
            }
            if(flag!=0)
            {
                printf ("Error input again\n",i);
                flag=0;
            }
            else
                flag=1;
        }
        for (i=0;i<4;i++)
            for (j=0;j<4;j++)
                if (j!=i)
                    for (k=0;k<4;k++)
                        if (k!=j && k!=i)
                            for (t=0;t<4;t++)
                                if (t!=i && t!=j && t!=k)
                                {
                                    res=treat(num[i],num[j],num[k],num[t]);
                                }
    if (res==0)
        printf ("\nNo answer\n");
    else ;
    //    printf ("time=%d\n\n",time);
    printf ("\n1: Go on\n2: Quit\n");
    scanf ("%d",&again);
    }
}

int treat(float a,float b,float c,float d)
{
    int i,j,k;
    float sum1,sum2,sum3;
    for (i=0;i<4;i++)
        for (j=0;j<6;j++)
            for (k=0;k<6;k++)
            {
                if ((!(i==3 && b==0)) && (!(j==3 && c==0)) && (!(k==3 && d==0)))
                {
                    sum1=myF(i,a,b);
                    sum2=myF(j,sum1,c);
                    sum3=myF(k,sum2,d);
                    if (fabs(sum3-24)<0.1)
                    {
                        temp++;
                        myPrint(1,i,j,k,a,b,c,d);
                    //    printf ("sum1:myF(%d,%2.0f,%2.0f)  sum1=%f\n",i,a,b,sum1);
                    //    printf ("sum2:myF(%d,%2.0f,%2.0f)  sum2=%f\n",j,c,d,sum2);
                    //    printf ("1:myF(%d,myF(%d,myF(%d,%2.0f,%2.0f),%2.0f),%2.0f)   sum3=%f\n\n",k,j,i,a,b,c,d,sum3);
                    }
                }
                if (k==2)
                {
                    sum1=myF(i,a,b);
                    sum2=myF(j,c,d);
                    sum3=sum1*sum2;
                     if (fabs(sum3-24)<0.1)
                    {
                        temp++;
                        myPrint(2,i,j,k,a,b,c,d);
                    //    printf ("sum1:myF(%d,%2.0f,%2.0f)  sum1=%f\n",i,a,b,sum1);
                    //    printf ("sum2:myF(%d,%2.0f,%2.0f)  sum2=%f\n",j,c,d,sum2);
                    //    printf ("2:myF(%d,myF(%d,%2.0f,%2.0f),myF(%d,%2.0f,%2.0f))   sum3=%f\n\n",k,i,a,b,j,c,d,sum3);
                    }
                }

                if (k==3)
                {
                    sum1=myF(i,a,b);
                    sum2=myF(j,c,d);
                    if (sum2!=0)
                    {
                        sum3=sum1/sum2;
                         if (fabs(sum3-24)<0.1)
                        {
                            temp++;
                            myPrint(3,i,j,k,a,b,c,d);
                        //    printf ("sum1:myF(%d,%2.0f,%2.0f)  sum1=%f\n",i,a,b,sum1);
                        //    printf ("sum2:myF(%d,%2.0f,%2.0f)  sum2=%f\n",j,c,d,sum2);
                        //    printf ("3:myF(%d,myF(%d,%2.0f,%2.0f),myF(%d,%2.0f,%2.0f))   sum3=%f\n\n",k,i,a,b,j,c,d,sum3);
                        }
                    }
                }
            }
    if (temp==0)
        return 0;
    else
        return 1;
}

float myF(int flag,float m,float n)
{    
//    time++;
    if (flag==0)
        return (m+n);
    if (flag==1)
        return (m-n);
    if (flag==2)
        return (m*n);
    if (flag==3)
        if (n==0)
            return 30000;
        else
            return (m/n);
    if (flag==4)
        return (n-m);
    if (flag==5)
        if (m==0)
            return 30000;
        else
            return (n/m);
    return 0;
}

void myPrint(int type,int i,int j,int k,float a,float b,float c,float d)
{
    char sigle[6];
        sigle[0]='+';
        sigle[1]='-';
        sigle[2]='*';
        sigle[3]='/';
        sigle[4]='-';
        sigle[5]='/';
    if (type==1){
        if(j==4 || j==5)
        {
            if (k==4 || k==5)
                printf("%2.0f %c (%2.0f %c (%2.0f %c %2.0f)) =24\n",d,sigle[k],c,sigle[j],a,sigle[i],b);
            else
                printf("(%2.0f %c (%2.0f %c %2.0f)) %c %2.0f =24\n",c,sigle[j],a,sigle[i],b,sigle[k],d);
        }
        else if (k==4 || k==5)
        {
            printf("%2.0f %c ((%2.0f %c %2.0f) %c %2.0f) =24\n",d,sigle[k],a,sigle[i],b,sigle[j],c);
        }
        else
            printf("((%2.0f %c %2.0f) %c %2.0f) %c %2.0f =24\n",a,sigle[i],b,sigle[j],c,sigle[k],d);
    }
    if (type==2 || type==3)
    {
    //    if (k==4 || k==5)
    //        printf("(%2.0f %c %2.0f) %c (%2.0f %c %2.0f)=24\n",c,sigle[j],d,sigle[k],a,sigle[i],b);
    //    else
            printf("(%2.0f %c %2.0f) %c (%2.0f %c %2.0f) =24\n",a,sigle[i],b,sigle[k],c,sigle[j],d);
    }
}                    

回复列表 (共57个回复)

沙发

Yes,7 Steps is enough!

板凳

我的意思是有没有可能用更少的步数?
路边看得的消息说外国有5步的神话

3 楼

强!

4 楼

平均5点几步还差不多!
5步一定猜出不可能!!!

5 楼

有算法吗?
我这两天老是想不下去,感觉要判断的太多,没有一个统一解决的办法,总不能if到底吧

6 楼

源代码 很烦
但算法很简单——穷举,把所有可能的表达式都算一遍
用了点小技巧:
我用两个表达式代表了可能的表达式:
((a※b)※c)※d
(a※b)※(c※d)
※代表“6”种运算符“+ - * /”和“--,//”
“--”和“//”是用来实现交换运算次序的
例如:“-”表示:m-n  “--”则表示:n-m;
      “/”表示:m/n   “//”表示: n/m
其余的不足可以由a,b,c,d任意交换顺序来实现(24种)
//在下面有一处 return 30000
//这里应该return:13*13*24+1;因为出现“除以0”的情况肯定不行,所以返回一个较大的数使结果>24

7 楼

5步之内猜中的概率可达到55.8%
6步之内猜中的概率可达到97.5%

8 楼

关于算“猜数字”思想
大学时写的程序遗失了,还好以前写的部分手稿找到了!
等周末我有空在来贴。

工作实在是忙啊,早7:00出发~晚7:00到家;辛苦啊!

9 楼

呵呵
忙了好啊,俺现在还是学校里的无产阶级

10 楼

那应该好好学习!!

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