Question

我正在编写一个模拟“左右中心”的程序。如果您不熟悉，则游戏涉及3个骰子，其中1个边为“ L”，1个边为“ R”，1个边为“ C”和3个边点。每个人都从$ 3开始。如果滚动“ L”，则向左传递一个美元。如果滚动“ R”，则向右传递一个美元。如果您按下“ C”，则将一个美元置于中间。如果滚动点，则不会采取任何措施。比赛然后转到左边。游戏继续进行，直到只有1个玩家有剩余钱，并且该玩家赢得了所有奖金。

除了一件奇怪的事情，我使程序正常运行。当我按以下方式运行时，它运行良好，通常需要70-150转才能完成。当我注释掉这些行

 cout << "In gameOver(). Number of brokeJamokes: " << brokeJamokes;
 cin.ignore();

该程序需要数十万（或数百万）次旋转才能完成。为什么简单的输出会改变呢？

完整代码如下：

#include <iostream>
#include <string>
#include <iomanip>

using namespace std;

int player[10] = {0};
int bank[10] = {0};


int rollDie() {

srand(time(NULL));

int randNum = rand()%6+1;

//cout << "In rollDie(), die roll is " << randNum << "\n";
//cin.ignore();

return randNum;
}

int distributeCash(int roll, int playerNum) {

if(roll == 1) { //pass left
    bank[playerNum]--;
    /* if active player is player 10 (player[9]), we need to pass to player 1 (player[0]) 
    instead of the nonexistant player 11, so we change the array value to -1 */
    if(playerNum == 9) {playerNum = -1; }

    bank[playerNum + 1]++;
    return 0;
}
 if(roll == 2) { //pass right
    bank[playerNum]--;
    /* if active player is player 1 (player[0]), we need to pass to player 10 (player[9]) 
    instead of the nonexistant player 0, so we change the array value to 11 */
    if(playerNum == 0) {playerNum = 10;}
    bank[playerNum - 1]++;
    return 0;
}
if(roll == 3) { //pass to center
    bank[playerNum]--;
    return 0;
}
else {
    return 0;
}
return 0;

}

int gameOver() {
int brokeJamokes = 0;

for(int i = 0; i < 10; i++) {
    if(bank[i] == 0) { brokeJamokes++; }
}

cout << "In gameOver(). Number of brokeJamokes: " << brokeJamokes;
cin.ignore();
if(brokeJamokes==9) {return 1;}
else return 0;

}

void showWinner() {
for(int i = 0; i < 10; i++) {
    if(bank[i] != 0) { 
        cout << "Player " << (i+1) << " is the winner!\n";
        cin.ignore();
        return;
    }
}
}
int main()
{

int roll[3] = {0};

for(int x = 1; x < 10; x++) { //initialize all banks to 3 except test player (player 1)
  bank[x] = 3;
}
bank[0] = 3;  //test player bank initialization
int turnCount = 0;

  while(!gameOver()){

  for(int i = 0; i < 10; i++) {
      if(gameOver()) {break;}
      for(int j = 0; j < 3; j++) {
         roll[j] = rollDie();
         if(bank[i] != 0) {
              distributeCash(roll[j], i);
          }
      }
     /* cout << "After player " << (i + 1) << "'s roll: \n";
      for(int l = 0; l < 10; l++) {
          cout << "Player " << (l + 1) << " $" << bank[l] << "\n";
      }
      cin.ignore();
  */

  turnCount++;}
  }
    showWinner();
    cout << "Number of turns: " << turnCount << "\n";

    cout << "Game over!\n";



}

Answer 1

正如melpomene所说，您反复称呼srand，并设置了相同的种子（因为我认为它使用时间需要第二种分辨率）。因此，直到时间改变，您将连续获得成千上万个具有相同值的“随机”数字。想想如果每个人都得到相同的掷骰，游戏将永远不会结束。

当您使用cout行时，它将大大降低程序速度，因此将srand设置为相同的值后，可以减少连续滚动的次数。

要解决此问题，请将对srand的调用移至主函数，以便仅调用一次。

注释掉cin.ignore会使程序运行大量的迭代

1 个答案: