C++ 中的石頭剪刀布

Question

我正在嘗試完成使用C++的編程原理和實踐練習10第4章中的一個練習。該練習即將編寫一個剪刀石頭布游戲，不使用隨機函數讓計算機在石頭剪刀布之間“選擇” ，所以你必須想辦法讓電腦選擇隨機。 順便說一句，這不是重點，重點是我的程序會跟蹤玩家和計算機得分。 現在的問題是我不知道如何分配分數。 我想出了這個：

enum types {ROCK, PAPER, SCISSORS}
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/* 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;

問題是我不認為這段代碼是好的。 有沒有更聰明的方法來做到這一點？

在互聯網上，我找到了數學家尼克麥克拉倫制作的版本。 這是他的代碼：

vector<string> words;
words.push_back("Rock");
words.push_back("Paper");
words.push_back("Scissors");

string guess;
cout << "Make a guess - Rock, Paper or Scissors\n";
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 used\n";
        yours = 0;
    }
    seed = (13*seed)%256;
    mine = seed%3;
    if (mine == yours) // draw
        cout << "Both guesses were " << words[yours] << " - no winner\n";
    else if ((3+mine-yours)%3 == 1) // computer wins
        cout << words[mine] << " beats " << words[yours] << " - I win\n";
    else // player wins
        cout << words[yours] << " beats " << words[mine] << " - you win\n";
}     

順便說一下， 這里是他制作的完整代碼的鏈接。 如果您查看整個代碼，也許您會了解更多。

所以他使用了另一種方法，但我不明白他的代碼的這一部分：

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

我知道如果該表達式的結果為 1，則計算機獲勝，否則獲勝者是玩家，但我不明白其中的邏輯。

Answer 1

你想看的是 mine-yours mod 3，如果 mine-yours == 1 mod 3 那么我的贏了，如果你不相信看這三種情況。

(3+mine-yours)%3

這着眼於差異 mod 3，+3 是為了確保 3+mine-yours 在被 % 評估之前是正數

如果您不確定 % https://www.cprogramming.com/tutorial/modulus.html

或者什么 mod 是https://en.wikipedia.org/wiki/Modular_arithmetic

Answer 2

這段代碼提醒我為什么聰明是善良的敵人。

我建議您查看此處發布的類似問題的答案： Rock, Paper, Scissors Game Java

基本上最好為選擇提供一個抽象類的接口，為不同的有效選擇提供實現，並讓類決定它是否從另一個實現中獲勝。

Answer 3

//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";
    }
}