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基本上，我需要找出如何在屏幕上输出值，以实现二进制搜索的成功部分。我试图更改的初始值，last但它要么崩溃，要么保持不变。我逐步浏览了代码，一切似乎都正常。

来自编译器的程序示例

#include<iostream>

using namespace std;

void binarySearch();

int main()
{
    //Called the function in main
    binarySearch();
    system("pause");
    return 0;
};

//void function for binary Search
void binarySearch()
{
      //creating array
      int array[100];
       array[0] = 1;
       int i,target;
      // using Boolean to create pattern for generated numbers in the array
      bool check = false;

      //loop to implement pattern
      for (int x = 1; x < 100; x++)
      {
        if (check == true)
        {
           array[x] = array[x - 1] + 1;
           check = false;
        }
        else
        {   
           array[x] = array[x - 1] + 2; 
           check = true;

        }
       }

     **Code found online and modified to fit**  

  int first,mid,last,completed,successful,tests;
  completed = 0;
  successful = 0;
  tests = 0;
  double percentage;
  percentage = 1;

  for(int x=0;x<100;x++)
  {
    // Initialize first and last variables.
    first = 0;
    last = 2;
    srand( (unsigned)time( NULL ) );      
    int target = (rand() % 150) + 1;

    while(first <= last)
    {
       mid = (first + last)/2;

       if(target > array[mid])
       {
          first = mid + 1;
          tests++;          
       }
       else if(target < array[mid])
       {
          last = mid + 1;
          tests++;
       }
       else
       {
          first = last - 1;
       }
       if(target == array[mid])
       {
          successful++;
       }
    }
    completed++;
  } 

**Area which the error occur No value for successful**
//Output on screen
cout << endl;
cout << "There were "<< completed <<" searches completed."<< endl; 
cout << "There were "<< successful <<" successful searches." << endl; 
cout << percentage <<"%"<<" of the searches were successful." << endl; 
cout << "There was an average of " << completed << " tests per search." << endl;
cout << endl;

}
4

2 回答 2

1

我想我会先将代码分解成更小的部分，我至少希望能够理解每一部分。我不够聪明，无法确定我理解和你一样大的“块” binarySearch。

它看起来像这样：

  int array[100];
   array[0] = 1;
   int i,target;
  // using Boolean to create pattern for generated numbers in the array
  bool check = false;

  //loop to implement pattern
  for (int x = 1; x < 100; x++)
  {
    if (check == true)
    {
       array[x] = array[x - 1] + 1;
       check = false;
    }
    else
    {   
       array[x] = array[x - 1] + 2; 
       check = true;

    }
   }

应该初始化一个数组，所以我把它放在一个单独的函数中，然后稍微简化一下。首先进入一个函数：

int init(int array[100]) {
   array[0] = 1;
   int i,target;
  // using Boolean to create pattern for generated numbers in the array
  bool check = false;

  //loop to implement pattern
  for (int x = 1; x < 100; x++)
  {
    if (check == true)
    {
       array[x] = array[x - 1] + 1;
       check = false;
    }
    else
    {   
       array[x] = array[x - 1] + 2; 
       check = true;

    }
   }
}

然后简化：

int init(int *array) { 
    array[0] = 1;

    for (int i=1; i<100; i++)
        array[i] = array[i-1] + 1 + (i&1);
}

然后我会对以下的二进制搜索部分做大致相同的事情binarySearch：

while(first <= last)
{
   mid = (first + last)/2;

   if(target > array[mid])
   {
      first = mid + 1;
      tests++;          
   }
   else if(target < array[mid])
   {
      last = mid + 1;
      tests++;
   }
   else
   {
      first = last - 1;
   }
   if(target == array[mid])
   {
      successful++;
   }

并简化：

int *binary_search(int const *left, int const *right, int *tests, int val) { 
    int const *mid = left + (right - left)/2;

    while (left < right) {
        ++*tests;
        if (val < *mid)
            right = mid;
        else
            left = mid + 1;
    }
    return left;
}

最后，您需要代码生成一个随机数、搜索它、跟踪搜索次数的统计信息以及成功的次数等。

于 2012-09-21T20:13:16.310 回答
0

你有几个问题。

首先，正如我评论的那样，您没有last正确初始化。

其次，在这种else if(target < array[mid])情况下，您没有last正确修改。您正在设置last = mid + 1;，但您需要改为设置last = mid - 1;

接下来，您没有正确退出循环。如果 first > last，则仅退出 while 循环，但应退出一次target == array[mid]；也许战略break;声明会有所帮助。

当我进行这三项更改时，应用程序每次都会运行直到完成，但是我会收到 0 次或 100 次成功的搜索——两者之间什么都没有。

不过，这应该足以让您走得更远。

于 2012-09-21T20:13:22.137 回答