输入任意四个整数(0到10)，运算符只有加减乘除，还有括号.每个数只能且必须用一次。要求判断这些表达的结果中是否有24。如果有，输出计算表达式：如输入4，6，1，1 输出 4*6*1*1 =24 （允许有括号）。
#include<stdio.h>
double Sum(double a,double b,int i)
{
    double sum;
    if(i==0)
        sum=a+b;
    else if(i==1)
        sum=a-b;
    else if(i==2)
        sum=a*b;
    else if(i==3)
    {
        if(b==0)
            return -1000000000000;
        else
            sum=a/b;
    }
    return sum;
}
int main()
{
    int i1,i2,i3,i4,j1,j2,j3,s[4];
    char a[4]={'+','-','*','/'};
    double s1,s2,s3;
    printf("请输入0-10之间的数：");
    scanf("%d%d%d%d",&s[0],&s[1],&s[2],&s[3]);
    printf("\n");
    for(i1=0;i1<4;i1++)
    {
        for(i2=0;i2<4;i2++)
        {

            if(i2!=i1)
            {
                for(j1=0;j1<4;j1++)
                {
                    s1=Sum(s[i1],s[i2],j1);
                    for(i3=0;i3<4;i3++)
                    {
                        if(i3!=i2&&i3!=i1)
                        {
                            for(j2=0;j2<4;j2++)
                            {
                                s2=Sum(s1,s[i3],j2);
                                for(i4=0;i4<4;i4++)
                                {
                                    if(i4!=i1&&i4!=i2&&i4!=i3)
                                    {
                                        for(j3=0;j3<4;j3++)
                                        {
                                            s3=Sum(s2,s[i4],j3);
                                            if(s3==24.0)
                                            {
                                                if((a[j1]=='+'||a[j1]=='-')&&(a[j2]=='*'||a[j2]=='/'))
                                                {printf("1q");
                                                    printf("(%d%c%d)%c%d%c%d=24\n",s[i1],a[j1],s[i2],a[j2],s[i3],a[j3],s[i4]);
                                                }
                                                else if((a[j2]=='+'||a[j2]=='-')&&(a[j3]=='*'||a[j3]=='/'))
                                                {printf("2q");
                                                    printf("(%d%c%d%c%d)%c%d=24\n",s[i1],a[j1],s[i2],a[j2],s[i3],a[j3],s[i4]);
                                                }
                                                else
                                                {printf("3q");
                                             printf("%d%c%d%c%d%c%d=24\n",s[i1],a[j1],s[i2],a[j2],s[i3],a[j3],s[i4]);
                                                }
                                       
                                            }
                                            if
                                        }
                                    }
                                }
                            }
                        }
                    }
                }
            }
        }
    }
    return 0;

}
                                

不会啊

例如6*8/(1+1)=24这样的式子时 括号在后面时 用语句输出时 总是晕晕的  给些提示就好

如果我接下来用if语句去判断1/24 ，能不能得出全部的答案？

感谢