java实现简单石头剪刀布游戏

  • Z时代
  • 2024-01-10
  • 分类：综合

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[0] == D[1] && D[0] != D[2]) {
          if ('0' == D[0] && '1' == D[2]) {
            A++;
            B++;
            C -= 2;
          }
          else if ('1' == D[0] && '2' == D[2]) {
            A++;
            B++;
            C -= 2;
          }
          else if ('2' == D[0] && '0' == D[2]) {
            A++;
            B++;
            C -= 2;
          }else {
            A--;
            B--;
            C += 2;
          }
        }
        // A and C success
 if (D[0] == D[2] && D[0] != D[1]) {
          if ('0' == D[0] && '1' == D[1]) {
            A++;
            B -= 2;
            C++;
          }
          else if ('1' == D[0] && '2' == D[1]) {
            A++;
            B -= 2;
            C++;
          }
          else if ('2' == D[0] && '0' == D[1]) {
            A++;
            B -= 2;
            C++;
          }else {
            A--;
            B += 2;
            C--;
          }
        }
        // C and B success
 if (D[1] == D[2] && D[1] != D[0]) {
          if ('0' == D[1] && '1' == D[0]) {
            A -= 2;
            B++;
            C++;
          }
          else if ('1' == D[1] && '2' == D[0]) {
            A -= 2;
            B++;
            C++;
          }
          else if ('2' == D[1] && '0' == D[0]) {
            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

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