Question

我正在制作一个简单的猜谜游戏程序。用户输入100之外的值，程序会告诉用户他们的猜测是高还是低。我想这样做，程序让用户知道他们之前已经输入了当前的号码。我如何在我的程序中实现一个循环，让用户猜测并将其与数组列表进行比较以查找重复值？

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

int main() 
{
  srand(time(0));
  const int SIZE = 100; //array
  int number[SIZE];
  int i;
  int numb = rand() % 100; //rng
  cout <<"Hint: "<< numb << endl;
  cout << " " << endl;
  cout << "I'm thinking of a number between 1 and 100. Guess what it is: ";
  for (i = 0; i < SIZE; i++)
  {
    cin >> number[i];
      if (number[i] == numb)  
      {
        cout << "Correct! It's " << numb << endl; //if user guesses correct
        break;
      }
      else if (number[i] < numb) //if user guesses too low
      {
        cout << "That's too low! guess again: " ;
      }
      else if (number[i] > numb) //if user guesses too high
      {
        cout <<  "That's too high! guess again: " ;
      }           
  }
}

Answer 1

为已输入的值创建一个向量：

vector <int> atyped;

创建一个检查并返回true或false的函数：

int check(int entered_value, vector <int> atyped) {
ret=0;
  for (int i=0;i<atyped.size();i++){
    if (enterd_value==atyped[i])
     ret=1;
  }
return ret;
}

然后你所要做的就是：

if (check(number[i])!=0)
cout << "Sorry, you already tried that one!" <<endl;

希望这至少能让您了解如何解决问题;）

Answer 2

使用std::find

for (i = 0; i < SIZE; i++)
{
  cin >> number[i];
  if  (number[i] == numb)
  {
    cout << "Correct! It's " << numb << endl; //if user guesses correct
    break;
  }
  else if (std::find(number, number + i, number[i]) != number + i) 
  {
    cout <<  "You already guessed that! guess again: " ;
  }
  else if  ... // existing higher or lower
}

指针范围[number, number + i)是以前的所有猜测。 find返回第一个匹配指针，如果没有匹配则返回结束（number + i）。

如何在数组中搜索先前已输入的int值

2 个答案: