java实现简单石头剪刀布游戏
本文实例为大家分享了java实现简单石头剪刀布游戏的具体代码，供大家参考，具体内容如下问题描述
Alice, Bob和Cindy一起玩猜拳的游戏。和两个人的猜拳类似,每一轮,他们会从石头、剪刀、布中各自选一个出拳,基本的胜负规则是石头赢剪刀、剪刀赢布、布赢石头。如果一轮中正好可以分成胜负两边,则负边的每个人要支付给胜边的每个人一块钱。如果无法分成胜负两边,则都不出钱。比如,如果Alice出石头,而Bob和Cindy都出布,则Alice要分支付Bob和Cindy一块钱。再如,如果Alice出石头, Bob出剪刀, Cindy出布,则都不出钱。他们三人共进行了n轮游戏,请问最后每个人净赚多少钱?即赚的钱减去支付的钱是多少?
代码
package Ring1270.pra.java01;
import java.util.Scanner;
/**
* finger-guessing game: * n:number of games * A: Person A's money * B: Person B's money * C: Person C's money * 0: Stand for stone * 1: Stand for Scissor * 2: Stand for cloth * rule1: Two persons give the same result means game over * Rule2: The money add 1 everytime which win * Rule3:The money less 1 everytime which fail * */public class D_FingerGuessingGame {
public static void main(String[] args) {
int A = 0;
int B = 0;
int C = 0;
Scanner scanner = new Scanner(System.in);
System.out.printf("The number of game:");
int n = scanner.nextInt();
StringBuffer stringBuffer = new StringBuffer();
for (int i = 0; i <= n; i++) {
String s = scanner.nextLine();
char[] D = s.toCharArray();
for (int j = 0; j < D.length; j++) {
//A and B success
if (D == D && D != D) {
if ('0' == D && '1' == D) {
A++;
B++;
C -= 2;
}
else if ('1' == D && '2' == D) {
A++;
B++;
C -= 2;
}
else if ('2' == D && '0' == D) {
A++;
B++;
C -= 2;
}else {
A--;
B--;
C += 2;
}
}
// A and C success
if (D == D && D != D) {
if ('0' == D && '1' == D) {
A++;
B -= 2;
C++;
}
else if ('1' == D && '2' == D) {
A++;
B -= 2;
C++;
}
else if ('2' == D && '0' == D) {
A++;
B -= 2;
C++;
}else {
A--;
B += 2;
C--;
}
}
// C and B success
if (D == D && D != D) {
if ('0' == D && '1' == D) {
A -= 2;
B++;
C++;
}
else if ('1' == D && '2' == D) {
A -= 2;
B++;
C++;
}
else if ('2' == D && '0' == D) {
A -= 2;
B++;
C++;
}
else {
A += 2;
B--;
C--;
}
}
break;
}
}
System.out.println(A);
System.out.println(B);
System.out.println(C);
}
}
运行截图
以上就是本文的全部内容，希望对大家的学习有所帮助，也希望大家多多支持CodeAE代码之家。
原文链接：https://blog.csdn.net/weixin_43388956/article/details/113938103
文档来源：服务器之家http://www.zzvips.com/article/183310.html
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