由 a*a + b*b = c*c
变形可得 a*a = (c-b) * (c+b)
因此遍历 c-b 就可以得到 b 和 c
比如 a=24
a*a = 576
576 可以是 8*72（ 此时b=(72-8)/2，c=(72+8)/2 ）
    可以是 6*96（ 此时b=(96-6)/2，c=(96+6)/2 ）
    可以是 4*144（ 此时b=(144-4)/2，c=(144+4)/2 ）
这个算法应该是最快的（当然，还可以继续优化，但主体算法不变）

#include <stdio.h>
#include <time.h>

#define N 10000 // 耗时0.140秒，总数12467

int main()
{
    int avoid_optimization = 0;
    clock_t t0 = clock();

    for( unsigned a=1; a<N; ++a )
    {
        for( unsigned c_minus_b=a-1; c_minus_b>0; --c_minus_b )
        {
            if( a*a % c_minus_b != 0 )
                continue;

            unsigned c_plus_b = a*a/c_minus_b;

            if( c_plus_b%2 != c_minus_b%2 )
                continue;

            unsigned b = (c_plus_b - c_minus_b)/2;
            unsigned c = (c_plus_b + c_minus_b)/2;

            if( a > b )
                continue;
            if( c >= N )
                break;

            ++avoid_optimization;
            //printf( "%2d^2 + %2d^2 = %2d^2\n", a, b, c );
        }
    }

    printf( "time = %.3f\ncount = %d\n", (clock()-t0+0.0)/CLOCKS_PER_SEC, avoid_optimization );
    return 0;
}