数据结构与算法之枚举（穷举）法 C++实现

#include<iostream>
const int COCKPR = 3;
const int HENPR = 5;
const int CHICKS = 3;  //原错误const double CHICKPR = 1/3; 1 3 为整形。1/3也为整形所以为0,又一次声明1钱能买3仅仅小鸡
void buyChicken(int money, int chooks);
int main()
{
    int money = 100;
    int chooks = 100;
    buyChicken(money, chooks);
    return 0;
}
void buyChicken(int money, int chooks)
{
    using namespace std;
    int MaxCock = money/COCKPR;
    int MaxHen = money/HENPR;
    int MaxChick = chooks;
	int MinChick = chooks-MaxCock-MaxHen;
    int cock,hen,chick;
    int count = 0;
    for(cock=0; cock<= MaxCock; cock++)
    {
        for(hen=0; hen<=MaxHen; hen++)
        {
            for(chick=MinChick; chick<=MaxChick; chick++)
            {
                if (0 == chick%3 && cock + hen + chick == chooks && COCKPR*cock + HENPR*hen + chick/CHICKS == money)
                    cout << "公鸡: " << cock << " 母鸡: " << hen << " 小鸡: " << chick << " 第 " << count << "有结果" <<endl;
                count ++;
            }
        }
    }
    cout << "总共枚举多少次：" << count <<endl;
}

#include<iostream>
int main()
{
    using namespace std;
    int t1,t2,t3,t4,t5;
    for (t1=1; t1<=9; t1++)
    {
        for(t2=0; t2<=9; t2++)
        {
            for(t3=0; t3<=9; t3++)
            {
                for(t4=0; t4<=9; t4++)
                {
                    for(t5=0; t5<=9; t5++)
                    {
                        if(t5*100000 + t5*10000 + t5*1000 + t5*100 + t5*10 + t5 == t5*t1 + t4*t1*10 + t3*t1*100 + t2*t1*1000 + t1 *t1*10000 )
                        {
                            cout << " 算的数值为: " << t1 << " 法的数值为: " << t2 << " 描的数值为: " <<t3 << " 述的数值为: " << t4 << " 题的数值为: " << t5<<endl;
                            cout << " " << t1 << " " << t2 << " " << t3 << " " << t4 << " " << t5 <<endl;
                            cout << "X " << t1<<endl;
                            cout << "________________\n";
                            cout << " " << t5 << " " << t5 << " " << t5 << " " << t5 << " " << t5 << " " << t5 <<endl;
                        }
                    }
                }
            }
        }
    }
    return 0;
}