Project Euler task 39

If p is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exactly three solutions for p = 120.

{20,48,52}, {24,45,51}, {30,40,50}

For which value of p 1000, is the number of solutions maximised?

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#include < iostream >

#include < math.h >

using namespace std;

int piSqr[500];

int main() {

double fSqrt22 = sqrt((double) 2) / ((double) 2);

for (int i = 0; i < 500; i++)

piSqr[i] = i * i;

int iMaxVariants = 0;

int iPMax = 0;

for (int p = 3; p <= 1000; p++) {

int iVariants = 0;

for (int c = 1; c < p / 2; c++)

for (int a = 1; a < (int) (fSqrt22 * (double) c); a++) {

int b = p - a - c;

if (piSqr[a] + piSqr[b] == piSqr[c])

iVariants++;

}

if (iVariants > iMaxVariants) {

iMaxVariants = iVariants;

iPMax = p;

}

}

cout << "p: " << iPMax << endl;

return 0;

}