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6.34题目

猜数字游戏)编写一个程序，可以玩“猜数字”的游戏。具体描述如下:程序在1~1000之间的整数中随机选择需要被猜的数，然后显示:

代码

#include <iostream>
#include <cstdlib>
#include <ctime>using namespace std;int main()
{srand(static_cast<unsigned int>(time(0)));while (1){int guessNumber = rand() % 1000 + 1;char YesOrNo;cout << "I have a number between 1 and 1000.\n";cout << "Can you guess my number ?\nPlease type your first guess.\n ";while (1){int guess;cin >> guess;if (guess > guessNumber){cout << "Too high.Try again.\n";}else if (guess < guessNumber){cout << "Too low.Try again.\n";}else{cout << "Excellent! You guessed the number!\n";cout << "Woule you like to play again (y or n)?\n";cin >> YesOrNo;break;}}if (YesOrNo == 'n')break;}return 0;
}

运行截图

6.35题目

(猜数字游戏的修改)修改练习题6.34 中的程序，统计玩家猜想的次数。如果次数没有超过10次打印“Either you know the secret or you got lucky!”。如果玩家10 次才猜中,打印出“Ahah! You knowthe secret!”。如果玩家超过10 次才猜中，打印“You should be able to do better!”。为什么猜测次数不应超过 10 次呢?因为在每次“好的猜想”过程中，玩家应该能够排除一半的数。现在说明了为什么任何1~1000之间的数字能够不超过10次就被猜中。

代码

#include <iostream>
#include <cstdlib>
#include <ctime>using namespace std;int main()
{srand(static_cast<unsigned int>(time(0)));while (1){int count = 0; // 用于计算猜测次数int guessNumber = rand() % 1000 + 1;char YesOrNo;cout << "I have a number between 1 and 1000.\n";cout << "Can you guess my number ?\nPlease type your first guess.\n ";while (1){int guess;cin >> guess;count++; // 猜测次数加一if (guess > guessNumber){cout << "Too high.Try again.\n";}else if (guess < guessNumber){cout << "Too low.Try again.\n";}else{cout << "Excellent! You guessed the number!\n";cout << "Woule you like to play again (y or n)?\n";cin >> YesOrNo;break;}}if (count < 10) // 添加的代码输出cout << "Either you know the secret or you got lucky!";else if (count == 10)cout << "Ahah! You know the secret!";elsecout << "You shoule be able to do better!";if (YesOrNo == 'n'){break;}}return 0;
}

运行截图