C語言版的算24游戲源碼

下邊代碼內(nèi)容是關(guān)于C語言版的算24游戲的代碼。

#include "stdafx.h"
#include <stdio.h>
#include <stdlib.h>

double sum(double a,double b)
{
    return a+b;
}
double mi(double a,double b)
{
    return a-b;
}
double mul(double a,double b)
{
}
double di(double a,double b)
{
    return a/b;
}

{
    int i;
    double t[4];
    for(i=0;i<4;i++)
        des[i]=src[i];
}

void reorder4(int cp4,double re4)
{
    switch(cp4)
    {
    case 0:
        num[0]=re4;
        num[1]=num[2];
        num[2]=num[3];
        break;
    case 1:
        num[1]=re4;
        num[2]=num[3];
        break;
    case 2:
        num[2]=re4;
        break;

    }
}

void reorder3(int cp3,double re3)
{
    switch(cp3)
    {
    case 0:
        num[0]=re3;
        num[1]=num[2];
        break;
    case 1:
        num[1]=re3;
        break;

    }
}

{
    if(opt==sum)
        return '+';
    else if(opt==mi)
        return '-';
    else if(opt==mul)
    else if(opt==di)
        return '/';

}

{
    double n0=num[0];
    double n1=num[1];
    double n2=num[2];
    double n3=num[3];

    switch(ta4)
    {
    case 0:
        if(ta3==0)
            printf("((%.0f%c%.0f)%c%.0f)%c%.0fn",n0,opc(op4),n1,opc(op3),n2,opc(op2),n3);
        else if(ta3==1)
            printf("(%.0f%c%.0f)%c(%.0f%c%.0f)n",n0,opc(op4),n1,opc(op2),n2,opc(op3),n3);
        break;
    case 1:
        if(ta3==0)
            printf("(%.0f%c(%.0f%c%.0f))%c%.0fn",n0,opc(op3),n1,opc(op4),n2,opc(op2),n3);
        else if(ta3==1)
            printf("%.0f%c((%.0f%c%.0f)%c%.0f)n",n0,opc(op2),n1,opc(op4),n2,opc(op3),n3);
        break;
    case 2:
        if(ta3==0)
            printf("(%.0f%c%.0f)%c(%.0f%c%.0f)n",n0,opc(op3),n1,opc(op2),n2,opc(op4),n3);
        else if(ta3==1)
            printf("%.0f%c(%.0f%c(%.0f%c%.0f))n",n0,opc(op2),n1,opc(op3),n2,opc(op4),n3);
        break;
    }
}

int main(void)
{

    scanf("%d%d%d%d",&a,&b,&c,&d);

    num[0]=(double)a;
    num[1]=(double)b;
    num[2]=(double)c;
    num[3]=(double)d;

    copyarr(num,numf);
    copyarr(num,num4);

    for(cp4=0;cp4<=2;cp4++)
    {
        for(i=0;i<4;i++)
        {
            cop4=ops[i];
            reorder4(cp4,re4);
            copyarr(num,num3);
            for(cp3=0;cp3<=1;cp3++)
            {
                for(j=0;j<4;j++)
                {
                    cop3=ops[j];
                    reorder3(cp3,re3);

                    for(k=0;k<4;k++)
                    {
                        cop2=ops[k];
                        if(re2==24)
                        {
                            printresult(numf,cp3,cp4,cop2,cop3,cop4);
                            system("pause");
                            return 0;
                        }
                    }
                    copyarr(num3,num);
                }
            }
            copyarr(num4,num);
        }
    }

    system("pause");
    return 0;
}