枚举法的本质就是从全部候选答案中去搜索正确的解,使用该算法须要满足两个条件:
1、能够先确定候选答案的数量。
2、候选答案的范围在求解之前必须是一个确定的集合。
枚举是最简单。最基础。也是最没效率的算法
枚举法长处:
1、枚举有超级无敌准确性。仅仅要时间足够。正确的枚举得出的结论是绝对正确的。
2、枚举拥有天下第一全面性,由于它是对全部方案的全面搜索。所以,它可以得出全部的解。
程序优化:
对于枚举算法,加强约束条件,缩小枚举的范围。是程序优化的主要考虑方向。
实例1:百钱买百鸡
一百个铜钱买了一百仅仅鸡。当中公鸡一仅仅3钱、母鸡一仅仅5钱。小鸡一钱3仅仅,问一百仅仅鸡中公鸡、母鸡、小鸡各多少)
一百个铜钱买了一百仅仅鸡。当中公鸡一仅仅3钱、母鸡一仅仅5钱。小鸡一钱3仅仅,问一百仅仅鸡中公鸡、母鸡、小鸡各多少)
代码:
#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 cock,hen,chick; int count = 0; for(cock=0; cock<= MaxCock; cock++) { for(hen=0; hen<=MaxHen; hen++) { for(chick=0; 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; }
执行结果:
仅仅对小鸡的数量加入一个最小的起始值
函数里面加 int MinChick = chooks-MaxCock-MaxHen;
把最里层小鸡数量的for循环的起始条件改为小鸡最小值 for(chick=MinChick; chick<=MaxChick; chick++)
运算时间复杂度例如以下。
时间复杂度降低一半。说明对于枚举算法。加强约束条件。缩小枚举的范围,是程序优化的主要考虑方向。
代码例如以下:
#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; }
实例2:填数字游戏
代码:
#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; }
执行结果: