c02p107a.txt

来自「Data Abstraction & Problem Solving with 」· 文本 代码 · 共 36 行

// Exercise 12

int search(int first, int last, int n);
int mystery(int n);

int main()
{
   cout << mystery(30) << endl;
   return 0;
}  // end main

int search(int first, int last, int n)
{
   int returnValue;

   cout << "Enter: first = " << first << " last = "
        << last << endl;

   int mid = (first + last)/2;
   if ( (mid * mid <= n) && (n < (mid+1) * (mid+1)) )
      returnValue = mid;
   else if (mid * mid > n)
      returnValue = search(first, mid-1, n);
   else
      returnValue = search(mid+1, last, n);

   cout << "Leave: first = " << first << " last = "
        << last << endl;
   return returnValue;
}  // end search

int mystery(int n)
{
   return search(1, n, n);
}  // end mystery