C++ 實現「撿樹枝」雙人博弈遊戲完整代碼

遊戲規則

• 樹枝總數 ≤ 21，用戶輸入。
• 擲硬幣決定先手。
• 每回合可取 1 或 2 根。
• 取走最後一根者輸（誤差遊戲變體）。

核心邏輯

採用模擬對戰，雙方輪流輸入取數，判斷勝負並統計多局勝率。

完整代碼

#include <iostream>
#include <ctime>
#include <cstdlib>
using namespace std;

int main() {
    char player[2][40];
    int rounds, totalSticks, currentPlayer, take, sticks;
    int win[2] = {0, 0};

    cout << "玩家1姓名: "; cin >> player[0];
    cout << "玩家2姓名: "; cin >> player[1];
    do { cout << "局數(1-5): "; cin >> rounds; } while (rounds < 1 || rounds > 5);
    do { cout << "每局樹枝數(1-21): "; cin >> totalSticks; } while (totalSticks < 1 || totalSticks > 21);

    for (int r = 1; r <= rounds; r++) {
        srand((unsigned)time(NULL) + r);
        currentPlayer = rand() % 2;
        sticks = totalSticks;
        cout << "\n第 " << r << " 局開始，共 " << sticks << " 根\n";

        while (sticks > 0) {
            cout << "剩餘: " << sticks << "  ";
            do { cout << player[currentPlayer] << " 取(1-2): "; cin >> take; } while (take < 1 || take > 2);
            sticks -= take;
            if (sticks <= 0) {
                cout << player[currentPlayer] << " 取走最後一根，輸了！\n";
                win[1 - currentPlayer]++; // 對方獲勝
                break;
            }
            currentPlayer = 1 - currentPlayer;
        }
    }
    cout << "\n最終戰績: " << player[0] << " " << win[0] << " 勝, " << player[1] << " " << win[1] << " 勝\n";
    return 0;
}

註

原題「誰拿到最後一根誰輸」屬於誤差遊戲，先手必勝策略為：保持每輪總取數為 3（對方取 1 我取 2，對方取 2 我取 1），使對方面對 1 根被迫取走。