本文實例為大家分享了java實現(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。和兩個人的猜拳類似,每一輪,他們會從石頭、剪刀、布中各自選一個出拳,基本的勝負規(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兩邊,則負邊的每個人要支付給勝邊的每個人一塊錢。如果無法分成勝負兩邊,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行了n輪游戲,請問最后每個人凈賺多少錢?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); }}

運行截圖

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