There exists exactly one Pythagorean triplet for which a + b + c = 1000. Find the product abc.
#include <iostream>
#include <limits>
#include <vector>
#include <math.h>
using namespace std;
void pause() { cin.ignore(numeric_limits<streamsize>::max(), '\n'); }
void practice1LoopTesting()
{
for (int i = 0; i < 5; i++)
{
cout << i << ' ';
}
}
void practice2LoopTesting()
{
for (int i = 0; i < 5; i++)
{
for (int j = 0; j < 5; j++)
{
cout << i << j << ' ';
}
}
}
void practice3LoopTesting()
{
for (int i = 0; i < 5; i++)
{
for (int j = 0; j < 5; j++)
{
for (int z = 0; z < 5; z++)
{
cout << i << j << z << ' ';
}
}
}
}
int main()
{
int a = 0;
int b = 0;
int c = 0;
int sumABC = 1000;
for (int a = 0; a < sumABC; a++)
{
for (int b = 0; b < sumABC; b++)
{
for (int c = 0; c < sumABC; c++)
{
// check 1
if (a < b && b < c)
{
// check 2
if (a*a + b*b == c*c)
{
// check 3
if (a + b + c == sumABC)
{
cout << a << ' ' << b << ' ' << c << ' ' << endl;
}
}
}
}
}
}
cout << "done";
// Practice with Combinations
/*cout << "Single for loop: " << endl;
practice1LoopTesting();
cout << "\n\n";
cout << "Double for loop: " << endl;
practice2LoopTesting();
cout << "\n\n";
cout << "Triple for loop: " << endl;
practice3LoopTesting();
cout << "\n\n";*/
pause();
}