C#使用二维数组模拟斗地主
本文实例讲述了C#使用二维数组模拟斗地主的方法。分享给大家供大家参考。具体如下：
packagecom.pb.demo;
importjava.util.Arrays;
importjava.util.Random;
/**
*扑克牌随机发牌♠♥♣♦二维数组实现
*
*/
publicclassPuker{
publicstaticvoidmain(String[]args){
//定义数组
String[][]puker=newString[5][];
puker[0]=newString[]{"♠A","♠2","♠3","♠4","♠5","♠6","♠7",
"♠8","♠9","♠10","♠J","♠Q","♠K"};
puker[1]=newString[]{"♥A","♥2","♥3","♥4","♥5","♥6","♥7",
"♥8","♥9","♥10","♥J","♥Q","♥K"};
puker[2]=newString[]{"♣A","♣2","♣3","♣4","♣5","♣6","♣7",
"♣8","♣9","♣10","♣J","♣Q","♣K"};
puker[3]=newString[]{"♦A","♦2","♦3","♦4","♦5","♦6","♦7",
"♦8","♦9","♦10","♦J","♦Q","♦K"};
puker[4]=newString[]{"大王","小王"};
//定义3个玩家和底牌
String[]player1=newString[17];
String[]player2=newString[17];
String[]player3=newString[17];
String[]temp=newString[3];
//二维数据洗牌
Randomrandom=newRandom();
inti1=0,i2=0,j1=0,j2=0;//定义４个变量，用来存放数组的下标
//洗牌1000次
for(inti=0;i<1000;i++){
//下标i1等于,随机0~4的整数一维的下标
i1=random.nextInt(5);//0~4之间
if(i1==4){
//如果一维的下标是4，则2维的的元素只有2个大王和小王,只有2个元素
j1=random.nextInt(2);
}else{
//如果不是4，则有13张牌
j1=random.nextInt(13);//0~12共13个元素
}
//因为最短的数组是2个元素,所以要2次，如果是多个就要多次
i2=random.nextInt(5);
if(i2==4){
j2=random.nextInt(2);
}else{
j2=random.nextInt(13);
}
//开始洗牌
Stringtmp=puker[i1][j1];
puker[i1][j1]=puker[i2][j2];//洗牌法，交换，打乱顺序
puker[i2][j2]=tmp;
}
//洗牌后的牌
System.out.println("===========洗牌后的顺序============");
for(inti=0;i<puker.length;i++){
for(intj=0;j<puker[i].length;j++){
System.out.print(puker[i][j]+"");
}
}
//开始发牌
for(inti=0;i<54;i++){
intp=i%3;//定义发给哪个玩家
intk=i/3-1;//定义轮次
if(i<3){
//先扣下三张底牌
temp[i]=puker[i/13][i%13];//从前向回取3张，估做底牌
}elseif(p==0){
player1[k]=puker[i/13][i%13];
}elseif(p==1){
player2[k]=puker[i/13][i%13];
}elseif(p==2){
player3[k]=puker[i/13][i%13];
}
}
System.out.println("\n==========发牌完成=============");
//玩家一
System.out.println("玩家一"+Arrays.toString(player1));
//玩家二
System.out.println("玩家二"+Arrays.toString(player2));
//玩家三
System.out.println("玩家三"+Arrays.toString(player3));
//底牌
System.out.println("底牌"+Arrays.toString(temp));
}
}
希望本文所述对大家的C#程序设计有所帮助。