C 中的岩石剪刀

Rock Paper Scissors in C++

本文关键字：石剪刀更新时间：2023-10-16

我正在尝试从书籍编程原理和练习中使用C 练习10第4章。，不使用随机功能让计算机在岩石，纸或剪刀之间"选择"，因此您必须找到一种使计算机选择随机的方法。顺便说一句，这不是重点，关键是我的程序跟踪播放器和计算机得分。现在问题是我不知道如何分配分数。我想到了：

enum types {ROCK, PAPER, SCISSORS}
.
.
.
.
.
/* this is the part where i check the score */
 // all cases the player would lose
if (player == ROCK && computer == PAPER)    // paper beats rock
    ++computer_score;
else if (player == PAPER && computer == SCISSORS)   // scissors beat paper
    ++computer_score;
else if (player == SCISSORS && computer == ROCK)    // rock beats scissors
    ++computer_score;
else // all other cases the player would win
    ++player_score;

问题是我认为这件代码不好。有什么更聪明的方法吗？

在互联网上，我发现了数学家尼克·麦克拉伦（Nick Maclaren）制作的版本。这是他的代码：

vector<string> words;
words.push_back("Rock");
words.push_back("Paper");
words.push_back("Scissors");
string guess;
cout << "Make a guess - Rock, Paper or Scissorsn";
while (cin >> guess) {
    int yours, mine; //-your- is the user choice. -mine- is the computer choice
    if (guess == "Rock")
        yours = 0;
    else if (guess  == "Paper")
        yours = 1;
    else if (guess  == "Scissors")
        yours = 2;
    else {
        cout << "Invalid guess - Rock usedn";
        yours = 0;
    }
    seed = (13*seed)%256;
    mine = seed%3;
    if (mine == yours) // draw
        cout << "Both guesses were " << words[yours] << " - no winnern";
    else if ((3+mine-yours)%3 == 1) // computer wins
        cout << words[mine] << " beats " << words[yours] << " - I winn";
    else // player wins
        cout << words[yours] << " beats " << words[mine] << " - you winn";
}

顺便说一句，这是他制作的完整代码的链接。如果您查看整个代码，也许您会了解更多。

所以他使用另一种方法，但我不了解他的代码的这一部分：

else if ((3+mine-yours)%3 == 1)//<-- I don't understand this conditional expression
    cout << words[mine] << " beats " << words[yours] <<
        " - I winn";

我得到的是，如果该表达式的结果是1，那么计算机赢了，否则获胜者是玩家，但我不了解逻辑。

您想看的是我的mod 3，如果我的== 1 mod 3，则我的胜利，如果您不相信这三种情况。

(3+mine-yours)%3

这介绍了差异mod 3， 3是要确保3 矿山是正面的，然后由％

评估

如果您不确定％https://www.cprogramming.com/tutorial/modulus.html

或什么是modhttps://en.wikipedia.org/wiki/modular_arithmetic

此代码提醒我为什么聪明是善良的敌人。

我建议您查看此处发布的类似问题的答案：摇滚，纸，剪刀游戏java

基本上，为选择提供抽象类的界面是很好的，为不同的有效选择提供实现，并让类决定它是否从其他实现中获胜。

//First time using the stack overflow, it just happend that I got a same 
//homework as the question,so here is my code for reference
#include <iostream>
#include <math.h>
#include <iomanip>
#include <time.h>
#include <string>
using namespace std;
//function prototype
int computerChoice();
int userChoice();
int result(int user, int computer);
string choiceName(int choice);
int main()
{
    //declare variable
    int user, computer, winner;
    //keep playing if there is a tie
    do {
        user = userChoice();
        computer = computerChoice();
        winner = result(user, computer);
        cout << "User choice : " << choiceName(user) << "tComputer Choice : " << choiceName(computer) << endl;
        if (winner == 1)
        {
            cout << "User win" << endl;
        }
        else if (winner == 2)
        {
            cout << "Computer win" << endl;
        }
        else
        {
            cout << "Tie" << endl;
        }
    } while (winner == 0);
    
    //exit system
    return 0;
}
//random generate a computer's choice 1-3
int computerChoice() {
    srand(time(0));
    return rand() % 3 + 1;
}
//get user input 1-3
int userChoice() {
    int userChoice;
    while (true)
    {
        cout << "Enter your choice" << endl;
        cout << "1.Rock" << endl;
        cout << "2.Paper" << endl;
        cout << "3.Scissors" << endl;
        cin >> userChoice;
        if (userChoice > 3 || userChoice < 1) {
            cout << "Invalid choice, try again" << endl;
            continue;
        }
        break;
    }
    return userChoice;
}
// 0 - tie      1- user win     2 - computer win
//rules
int result(int userChoice, int computerChoice) {
    if (userChoice == computerChoice) 
    {
        return 0;
    }
    else if (userChoice == 3 || computerChoice == 3) 
    {
        return userChoice < computerChoice ? 1 : 2;
    }
    else 
    {
        return userChoice > computerChoice ? 1 : 2;
    }
}
//match the choice name
string choiceName(int choice) {
    if (choice == 1) 
    {
        return "Rock";
    }
    else if (choice == 2) 
    {
        return "Paper";
    }
    else 
    {   
        return "Scissors";
    }
}