Hnu 12886 cracking the safe (brute force enumeration)

Question link: http://acm.hnu.cn/online? Action = problem & type = show & id = 12886> courseid = 274

Solution Report: enter four numbers. You can use parentheses to determine whether a formula with the +,-, *,/, and four arithmetic operations can produce a result of 24.

Explain the fourth example of the test: it should be 6/(1-3/4)

What are the three types of symbols in the brute force enumeration mode? Then, the sequence of these three symbol operations is enumerated, and then the 24 arrangement modes of the four numbers are enumerated. The time is 4 ^ 3*6*24.

Then pay attention to use the double type to determine whether it is equal to 24. FABS (ANS-24.0) <= 0.000000001

 1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 #include<map> 6 #include<string> 7 #include<cmath> 8 using namespace std; 9 int fu[4];10 int mei[50][5] = {11 {1 ,2 ,3 ,4},12 {1 ,2 ,4 ,3},13 {1 ,3 ,2 ,4},14 {1 ,3 ,4 ,2},15 {1 ,4 ,2 ,3},16 {1 ,4 ,3 ,2},17 {2 ,1 ,3 ,4},18 {2 ,1 ,4 ,3},19 {2 ,3 ,1 ,4},20 {2 ,3 ,4 ,1},21 {2 ,4 ,1 ,3},22 {2 ,4 ,3 ,1},23 {3 ,1 ,2 ,4},24 {3 ,1 ,4 ,2},25 {3 ,2 ,1 ,4},26 {3 ,2 ,4 ,1},27 {3 ,4 ,1 ,2},28 {3 ,4 ,2 ,1},29 {4 ,1 ,2 ,3},30 {4 ,1 ,3 ,2},31 {4 ,2 ,1 ,3},32 {4 ,2 ,3 ,1},33 {4 ,3 ,1 ,2},34 {4 ,3 ,2 ,1},35 };36 double CC(int f,double a,double b)37 {38     if(f == 4 && b == 0) return 0;39     if(f == 1) return a + b;40     if(f == 2) return a - b;41     if(f == 3) return a * b;42     if(f == 4) return a / b;43 }44 double calc(int l,double a,double b,double c,double d)45 {46 //    double a = num[1],b = num[2],c = num[3],d = num[4];47     if(l == 1)48     return CC(fu[3],CC(fu[2],CC(fu[1],a,b),c),d);49     if(l == 2)50     return CC(fu[2],CC(fu[1],a,b),CC(fu[3],c,d));51     if(l == 3)52     return CC(fu[3],CC(fu[1],a,CC(fu[2],b,c)),d);53     if(l == 4)54     return CC(fu[1],a,CC(fu[3],CC(fu[2],b,c),d));55     if(l == 5)56     return CC(fu[2],CC(fu[1],a,b),CC(fu[3],c,d));57     if(l == 6)58     return CC(fu[1],a,CC(fu[2],b,CC(fu[3],c,d)));59 }60 int main()61 {62     63     int T;64     scanf("%d",&T);65     while(T--)66     {67         double num[5];68         scanf("%lf%lf%lf%lf",&num[1],&num[2],&num[3],&num[4]);69         int flag = 1;70         for(int i = 1;flag && i <= 4;++i)71         for(int j = 1;flag && j <= 4;++j)72         for(int k = 1;flag && k <= 4;++k)73         {74             fu[1] = i;75             fu[2] = j;76             fu[3] = k;77             for(int m = 0;flag && m < 24;++m)78             {79                 for(int l = 1;flag && l <= 6;++l)80                 {81                     double ans = calc(l,num[mei[m][0]],num[mei[m][1]],num[mei[m][2]],num[mei[m][3]]);82                     if(fabs(ans-24.0) <= 0.00000001)83                     {84                         flag = 0;85                         break;86                     }87                 }88             }89         }90         printf(flag? "NO\n":"YES\n");91     }92     return 0;93 }

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