HELP

[cash]

How WOULD I CHANGE THIS CODE SO THAT THE PROGRAM FINDS THE SMALLEST INTEGER THAT CAN BE EXPRESSED AS THE SUM OF SQUARES OF TWO DIFFERENT PATHS?

  #include <iostream>
using namespace std;


void sumSquare(int n)
{
    int i,j =1;
    for (i = 1; i * i <= n; i++)
        for ( j = 1; j * j <= n; j++)
            if (i * i + j * j == n) {
                cout << i << "^2 + "
                << j << "^2 = " << n<< endl;
            }
}


int main()
{


    int n = 50;
    sumSquare(n);
}

[Grime]

You need to find the smallest integer that can be expressed as the sum of square of two numbers, okay, where are the two numbers?

Get number 'A'.
Get number 'B'.

Required = A*A + B*B.
Yes?

To find the two smallest numbers whose square equals the given number, you can something like this:

void sumSquare(int n)
{
    int lowesti, lowestj;
    bool is_found = false;
    int i, j = 1;

    for (i = 1; i * i <= n; i++)
        for (j = 1; j * j <= n; j++)
            if (i * i + j * j == n) {
                if (!is_found) {
                    lowesti = i;
                    lowestj = j;
                    is_found = true;
                }
                else {
                    if (i + j < lowesti + lowestj) {
                        lowesti = i;
                        lowestj = j;
                    }
                }
            }
}

[cash]

the program should find the smallest number n= A*A + B*B= C*C + D*D

[Grime]

Ah okay.

So do the same thing but check for lowesti + lowestj == i+j later (once you've found lowesti and lowestj). And if nothing is found then do the first loop again but make sure you don't take the same values for lowesti and lowestj.

Donno if there's a better way.

[cash]

can you show me?

[Grime]

Give it a try

[Ganado]

Loop a number, n, from a small number, either until you find a match, or until a sufficiently large number.

Change sumSquare to return a bool, and have it keep track of the number of times the pattern you're looking for (A*A + B*B == n) is found. If it's found twice, and (A*A != C*C and A*A != D*D), then return true. Otherwise, if you've exhausted all the choices, return false and try the next highest n.

[salem c]

HELP isn't a topic title.
http://www.catb.org/esr/faqs/smart-questions.html#bespecific

Running around screaming in ALL CAPS isn't helpful either.
http://www.catb.org/esr/faqs/smart-questions.html#writewell

[lastchance]

#include <iostream>
#include <set>
#include <algorithm>
using namespace std;

int main()
{
   set<int> S;
   int d = 0, n = 10000000;
   bool best = false;

   while( !best )                 // Seek smallest n with c <= d and c^2+d^2 seen before
   {
      best = true;
      for ( ++d; best && d * d < n; ++d )
      {
         int dd = d * d;
         int ccmax = min( n - 1 - dd, dd );      // Seek to reduce current n with c <= d
         for ( int c = 1; best && c * c <= ccmax; c++ )
         {
            int test = dd + c * c;
            if ( !( S.insert( test ).second ) )  // Seen before
            {
               n = test;
               best = false;                     // Get out of c and d loops
            }
         }
      }
   }

   cout << n;
}

or even

#include <iostream>
#include <cmath>
using namespace std;

int main()
{
   for ( int n = 2; ; n++ )
   {
      bool found = false;
      int mid = sqrt( n / 2.0 );
      for ( int d = mid; d * d < n; ++d )
      {
         for ( int c = 1; c <= mid ; c++ )
         {
            if ( c * c + d * d == n )
            {
               if ( found )
               {
                  cout << n;
                  return 0;
               }
               found = true;
            }
         }
      }
   }
}