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Merge branch 'project_euler/problem_09' into project_euler/master

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* project_euler/problem_09:
  updating DIRECTORY.md
  optimized solution using only one loop copied from - https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_09/sol2.py
  updating DIRECTORY.md
  brute force method for Euler# 09

# Conflicts:
#	DIRECTORY.md
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* Problem 08
* [Sol1](https://github.com/TheAlgorithms/C/blob/master/project_euler/Problem%2008/sol1.c)
* [Sol2](https://github.com/TheAlgorithms/C/blob/master/project_euler/Problem%2008/sol2.c)
* Problem 09
* [Sol1](https://github.com/TheAlgorithms/C/blob/master/project_euler/Problem%2009/sol1.c)
* [Sol2](https://github.com/TheAlgorithms/C/blob/master/project_euler/Problem%2009/sol2.c)
## Searching
* [Binary Search](https://github.com/TheAlgorithms/C/blob/master/searching/Binary_Search.c)

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#include <stdio.h>
int main(void)
{
for (int a = 1; a < 300; a++)
for (int b = a+1; b < 400; b++)
for (int c = b+1; c < 500; c++)
{
if (a * a + b * b == c * c)
if (a + b + c == 1000)
{
printf("%d x %d x %d = %ld\n", a, b, c, (long int) a*b*c);
return 0;
}
}
return 0;
}

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#include <stdio.h>
#include <stdlib.h>
/**
Problem Statement:
A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
a^2 + b^2 = c^2
For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
Given a^2 + b^2 = c^2 and a+b+c = n, we can write:
b = (n^2 - 2*a*n) / (2*n - 2*a)
c = n - a - b
**/
int main(void)
{
int N = 1000;
for (int a = 1; a < 300; a++)
{
long tmp1 = N * N - 2 * a * N;
long tmp2 = 2 * (N - a);
div_t tmp3 = div(tmp1, tmp2);
int b = tmp3.quot;
int c = N - a - b;
if (a * a + b * b == c * c)
{
printf("%d x %d x %d = %ld\n", a, b, c, (long int) a*b*c);
return 0;
}
}
return 0;
}