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java實(shí)現(xiàn)簡單石頭剪刀布游戲

更新時(shí)間：2021年02月23日 14:10:27 作者：MTing•cat

這篇文章主要介紹了java實(shí)現(xiàn)簡單石頭剪刀布游戲，文中示例代碼介紹的非常詳細(xì)，具有一定的參考價(jià)值，感興趣的小伙伴們可以參考一下

本文實(shí)例為大家分享了java實(shí)現(xiàn)簡單石頭剪刀布游戲的具體代碼，供大家參考，具體內(nèi)容如下

問題描述

Alice, Bob和Cindy一起玩猜拳的游戲。和兩個(gè)人的猜拳類似,每一輪,他們會從石頭、剪刀、布中各自選一個(gè)出拳,基本的勝負(fù)規(guī)則是石頭贏剪刀、剪刀贏布、布贏石頭。如果一輪中正好可以分成勝負(fù)兩邊,則負(fù)邊的每個(gè)人要支付給勝邊的每個(gè)人一塊錢。如果無法分成勝負(fù)兩邊,則都不出錢。比如,如果Alice出石頭,而Bob和Cindy都出布,則Alice要分支付Bob和Cindy一塊錢。再如,如果Alice出石頭, Bob出剪刀, Cindy出布,則都不出錢。他們?nèi)斯策M(jìn)行了n輪游戲,請問最后每個(gè)人凈賺多少錢?即賺的錢減去支付的錢是多少?

代碼

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);
  }
}

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