C++ :

#include <iostream>
#include<cmath>
using namespace std;

int main() {
    int n;
    cin >> n;

    int cnt = 0;
    for (int c = 1; c <= n; ++c) {
        int c2 = c * c;
        for (int a = 1; a <= c; ++a) {
            int a2 = a * a;
            // 由 a<=b<=c，可推出 b 的范围
            int b2 = c2 - a2;
            if (b2 < a2) break;   // 此时 b<a，不符合 a<=b
            int b = static_cast<int>(sqrt(b2) + 0.5); // 四舍五入取整
            if (a <= b && b <= c && a * a + b * b == c * c) {
                ++cnt;
            }
        }
    }
    cout << cnt << endl;
    return 0;
}