Hy,我想要计算玩家一方面可以获得的所有扑克牌组合并显示所有组合。我不关心有多少双,满屋等。我只想计算玩家可以得到的所有可能的牌。因此,一只手由5张牌组成,必须有4种颜色(套装),一套牌必须重复。最大值是4个数字相同,第5张卡的数量必须不同。所有可能的手组合的正确结果必须是2598960(52以上5是2598960)。我只需要使用我的代码得到正确的结果,但我不知道如何编写算法
我有card
。cs class:
class card
{
private int number;
private char suit;
public card(int _number, char _suit)
{
this.number = _number;
this.suit = _suit;
}
public int Number
{
get{return this.number;}
set{this.number = value;}
}
public char Suit
{
get { return this.suit; }
set { this.suit = value; }
}
}
并且在Program
。cs类中,我有Main和这段代码:
static void Main(string[] args)
{
char[] suits = { '\u2660', '\u2663', '\u2665', '\u2666' }; //Spades, Clubs, Hearts, Diamonds
int[] numbers = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 };
int cnt = 0;
List<card> deck = new List<card>();
//making deck
foreach (char suit in suits)
{
foreach (int number in numbers)
{
deck.Add(new card(number, suit));
}
}
foreach (card first in deck)
{
foreach (card second in deck)
{
if (second.Number != first.Number)
{
foreach (card third in deck)
{
if (third.Number != second.Number && third.Number != first.Number)
{
foreach (card fourth in deck)
{
if (fourth.Number != third.Number && fourth.Number != second.Number && fourth.Number != first.Number)
{
foreach (card fifth in deck)
{
if (fifth.Suit != first.Suit && fifth.Suit != second.Suit && fifth.Suit != third.Suit && fifth.Suit != fourth.Suit)
{
//Console.WriteLine("{0}{1} {2}{3} {4}{5} {6}{7} {8}{9}", first.Number, first.Suit, second.Number, second.Suit, third.Number, third.Suit, fourth.Number, fourth.Suit, fifth.Number, fifth.Suit);
cnt++;
}
}
}
}
}
}
}
}
}
Console.WriteLine("Combinations: {0}", cnt);//Result must be: 2598960
}
答案 0 :(得分:3)
好吧,尽管我很欣赏你正在做的事情,但我觉得这更容易阅读:
int count = 0;
int cards_amount = 52;
for (var first = 0; first < cards_amount-4; first++)
for (var second = first + 1; second < cards_amount-3; second++)
for (var third = second+1; third < cards_amount-2; third++)
for (var fourth = third+1; fourth < cards_amount-1; fourth++)
for (var fifth = fourth+1; fifth < cards_amount; fifth++)
count++;
而不是查看所有重复项和if this card is different from the previous
,我所做的是:将所有卡片连成一行。选择第一个,然后选择剩下的第二个,然后选择剩下的第三个......依此类推。
这样,您无需检查倍数,并得到正确答案:)
修改强>
至于评论......只需在初始化列表后添加此行:
var deck_array = deck.ToArray();
您可能希望在class card
上执行此操作:
public override string ToString() {
// have this print your card number & suit
}
然后,只需更改count++;
行:
{
count ++;
Console.WriteLine("{0},{1},{2},{3},{4}", deck_array[first], deck_array[second],
deck_array[third] , deck_array[fourth] , deck_array[fifth] );
}
解决了......(现在你的卡片知道如何打印自己,而你最后只是打印了手。