Algorithms

for (i=0 ; i < iMax-1; i++)
{
  int temp = a[i];
  if (temp > 51) continue;

  for ( j=i+1; j < iMax; j++)
  {
    if (temp + a[j] == 51)
    {
      // do whatever you need...
    }
  }
}

LongHC wrote:

Imagine you have a array of integers, and you want to get any two numbers whose sum is 51, what you'll do?

int x = 11;
int y = 0;
 
while(x > y)
{
    x--;
    y++;   
}

Sort(array);
 
List list;
int x = 11;
int y = 0;
 
while(x > y)
{
    int index = BinarySearch(array, x);
 
    if(index >= 0)
    {
        list.Add(index);
    }
 
    index = BinarySearch(array, y);
 
    if(index >= 0)
    {
        list.Add(index);
    }
 
    x--;
    y++;   
}

cp9876 wrote:

If the integers are positive, then I can think of an O(n) solution.

cp9876 wrote:

As this sounds like homework

for (int i = 0 ; i < 51 ; i++)
   {
   if (a[i] < 51)
      {
      int n = 51 - a[i];
      if (c[n] >= 0)
        {
        // found one,  at a[i], a[c[n]]
        } 
      else
        {
        c[n] = i;
        }
      }
   }