03凑算式（数字填空）

//先给出答案：29
//有两种思路：1、暴力枚举（这里讲这个）  2、全排列（下次补上）
//注意啦：第一：浮点数注意事项。要不就使分子*1.0，然后abs(等式左边-等式右边)<1E-10;要不就左右两边通分，使分式运算化成整式运算。
//第二，关于“DEF”、"GHI"。这里DEF=D*100+E*10+F(而不是D*E*F,小心陷阱)
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
    int A, B, C, D, E, F, G, H, I;
    int count = 0;
    for (A = 1; A <= 9; A++)
    {
        for (B = 1; B <= 9; B++)
        {
            if (A == B)continue;
            for (C = 1; C <= 9; C++)
            {
                if (A == C || B == C)continue;
                for (D = 1; D <= 9; D++)
                {
                    if (A == D || B == D || C == D)continue;
                    for (E = 1; E <= 9; E++)
                    {
                        if (A == E || B == E || C == E || D == E)continue;
                        for (F = 1; F <= 9; F++)
                        {
                            if (A == F || B == F || C == F || D == F || E == F)continue;
                            for (G = 1; G <= 9; G++)
                            {
                                if (A == G || B == G || C == G || D == G || E == G || F == G)continue;
                                for (H = 1; H <= 9; H++)
                                {
                                    if (A == H || B == H || C == H || D == H || E == H || F == H || G == H)continue;
                                    for (I = 1; I <= 9; I++)
                                    {
                                        if (A == I || B == I || C == I || D == I || E == I || F == I || G == I || H == I)continue;
                                        int DEF = D * 100 + E * 10 + F;
                                        int GHI = G * 100 + H * 10 + I;
                                        if (abs(A + B*1.0 / C + DEF*1.0 / GHI-10)<1E-10)
                                        {
                                            count++;
                                        }
                                    }
                                }
                            }
                        }
                    }
                }
            }
        }
    }
    cout << count << endl;
}