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https://github.com/TheAlgorithms/C
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42 lines
1011 B
C
42 lines
1011 B
C
/**
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* \file
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* \brief [Problem 9](https://projecteuler.net/problem=9) solution
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*
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Problem Statement:
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A Pythagorean triplet is a set of three natural numbers, \f$a < b < c\f$,
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for which, \f$a^2 + b^2 = c^2\f$. For example, \f$3^2 + 4^2 = 9 + 16 = 25 =
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5^2\f$. There exists exactly one Pythagorean triplet for which \f$a + b + c =
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1000\f$. Find the product abc.
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Given \f$a^2 + b^2 = c^2\f$ and \f$a+b+c = n\f$, we can write:
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\f{eqnarray*}{
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b &=& \frac{n^2 - 2an}{2n - 2a}\\
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c &=& n - a - b
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\f}
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*/
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#include <stdio.h>
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#include <stdlib.h>
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/** Main function */
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int main(void)
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{
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int N = 1000;
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for (int a = 1; a < 300; a++)
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{
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long tmp1 = N * N - 2 * a * N;
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long tmp2 = 2 * (N - a);
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div_t tmp3 = div(tmp1, tmp2);
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int b = tmp3.quot;
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int c = N - a - b;
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if (a * a + b * b == c * c)
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{
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printf("%d x %d x %d = %ld\n", a, b, c, (long int)a * b * c);
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return 0;
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}
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}
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return 0;
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} |