(Problem 39)Integer right triangles

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?

题目大意：

如果p是一个直角三角形的周长，三角形的三边长{a,b,c}都是整数。对于p = 120一共有三组解：

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

对于1000以下的p中，哪一个能够产生最多的解？

#include <stdio.h>



int count(int p)

{

    int a, b, c, n, p1, p2;

    n = 0;

    p1 = p / 3;

    p2 = p / 2;

    for(a = 1; a < p1; a++) {

        for(b = p2 - a; b < p2; b++) {

            c = p - a - b;

            if(a * a + b * b == c * c) {

                n++;

            }

        }

    }    

    return n;

}



int main()

{

    int i, max, t, k;

    max = 3;

    for(i = 120; i < 1000; i++) {

        t = count(i);

        if(t > max) {

            max = t;

            k = i;

        }

    }

    printf("%d\n",k);

    return 0;

}

Answer:

840