石头剪子布

- - C语言实现
  - C++实现
  - Java实现
  - Python实现

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石头剪子布，是一种猜拳游戏。起源于中国，然后传到日本、朝鲜等地，随着亚欧贸易的不断发展它传到了欧洲，到了近现代逐渐风靡世界。简单明了的规则，使得石头剪子布没有任何规则漏洞可钻，单次玩法比拼运气，多回合玩法比拼心理博弈，使得石头剪子布这个古老的游戏同时用于“意外”与“技术”两种特性，深受世界人民喜爱。

游戏规则：石头打剪刀，布包石头，剪刀剪布。
现在，需要你写一个程序来判断石头剪子布游戏的结果。

输入

输入包括N+1行：
第一行是一个整数N，表示一共进行了N次游戏。1 <= N <= 100。
接下来N行的每一行包括两个字符串，表示游戏参与者Player1，Player2的选择（石头、剪子或者是布）：
S1 S2
字符串之间以空格隔开S1,S2只可能取值在{“Rock”, “Scissors”, “Paper”}（大小写敏感）中。

输出

输出包括N行，每一行对应一个胜利者（Player1或者Player2），或者游戏出现平局，则输出Tie。

样例输入

3
Rock Scissors
Paper Paper
Rock Paper

样例输出

Player1
Tie
Player2

提示
Rock是石头，Scissors是剪刀，Paper是布。

C语言实现

#include <stdio.h>
#include <string.h>

#define MAX_N 100

int main() {
int n;
scanf(“%d”, &n); // 读取游戏进行的次数N

char player1[MAX_N][10];
char player2[MAX_N][10];for (int i = 0; i < n; i++) {scanf("%s %s", player1[i], player2[i]);  // 读取每一轮两个玩家的选择
}for (int i = 0; i < n; i++) {if (strcmp(player1[i], "Rock") == 0 && strcmp(player2[i], "Scissors") == 0) {printf("Player1\n");  // 石头赢剪刀} else if (strcmp(player1[i], "Scissors") == 0 && strcmp(player2[i], "Paper") == 0) {printf("Player1\n");  // 剪刀赢布} else if (strcmp(player1[i], "Paper") == 0 && strcmp(player2[i], "Rock") == 0) {printf("Player1\n");  // 布赢石头} else if (strcmp(player1[i], "Scissors") == 0 && strcmp(player2[i], "Rock") == 0) {printf("Player2\n");  // 剪刀输石头} else if (strcmp(player1[i], "Paper") == 0 && strcmp(player2[i], "Scissors") == 0) {printf("Player2\n");  // 布输剪刀} else if (strcmp(player1[i], "Rock") == 0 && strcmp(player2[i], "Paper") == 0) {printf("Player2\n");  // 石头输布} else {printf("Tie\n");  // 平局情况}
}return 0;

}

C++实现

#include <iostream>
#include <string>
using namespace std;

int main() {
int n;
cin >> n; // 读取游戏进行的次数N

string player1[100];
string player2[100];for (int i = 0; i < n; i++) {cin >> player1[i] >> player2[i];  // 读取每一轮两个玩家的选择
}for (int i = 0; i < n; i++) {if (player1[i] == "Rock" && player2[i] == "Scissors") {cout << "Player1\n";  // 石头赢剪刀} else if (player1[i] == "Scissors" && player2[i] == "Paper") {cout << "Player1\n";  // 剪刀赢布} else if (player1[i] == "Paper" && player2[i] == "Rock") {cout << "Player1\n";  // 布赢石头} else if (player1[i] == "Scissors" && player2[i] == "Rock") {cout << "Player2\n";  // 剪刀输石头} else if (player1[i] == "Paper" && player2[i] == "Scissors") {cout << "Player2\n";  // 布输剪刀} else if (player1[i] == "Rock" && player2[i] == "Paper") {cout << "Player2\n";  // 石头输布} else {cout << "Tie\n";  // 平局情况}
}return 0;

}

Java实现

import java.util.Scanner;

public class RockPaperScissors {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt(); // 读取游戏进行的次数N

    String[] player1 = new String[n];String[] player2 = new String[n];for (int i = 0; i < n; i++) {player1[i] = scanner.next();player2[i] = scanner.next();  // 读取每一轮两个玩家的选择}for (int i = 0; i < n; i++) {if ("Rock".equals(player1[i]) && "Scissors".equals(player2[i])) {System.out.println("Player1");  // 石头赢剪刀} else if ("Scissors".equals(player1[i]) && "Paper".equals(player2[i])) {System.out.println("Player1");  // 剪刀赢布} else if ("Paper".equals(player1[i]) && "Rock".equals(player2[i])) {System.out.println("Player1");  // 布赢石头} else if ("Scissors".equals(player1[i]) && "Rock".equals(player2[i])) {System.out.println("Player2");  // 剪刀输石头} else if ("Paper".equals(player1[i]) && "Scissors".equals(player2[i])) {System.out.println("Player2");  // 布输剪刀} else if ("Rock".equals(player1[i]) && "Paper".equals(player2[i])) {System.out.println("Player2");  // 石头输布} else {System.out.println("Tie");  // 平局情况}}scanner.close();
}

}

Python实现

n = int(input())  # 读取游戏进行的次数Nfor _ in range(n):player1, player2 = input().split()  # 读取每一轮两个玩家的选择if player1 == "Rock" and player2 == "Scissors":print("Player1")  # 石头赢剪刀elif player1 == "Scissors" and player2 == "Paper":print("Player1")  # 剪刀赢布elif player1 == "Paper" and player2 == "Rock":print("Player1")  # 布赢石头elif player1 == "Scissors" and player2 == "Rock":print("Player2")  # 剪刀输石头elif player1 == "Paper" and player2 == "Scissors":print("Player2")  # 布输剪刀elif player1 == "Rock" and player2 == "Paper":print("Player2")  # 石头输布else:print("Tie")  # 平局情况