Atcoder解いてみた ABC 133 B - Good Distance

#ABC 133 B - Good Distance 解答例
問題はこちら

#include <iostream>
#include <string>
#include <map>
#include <cstring>
#include <cmath>
using namespace std;

int main() {
    int N; // 点の数2<=, <=10
    int D; // 次元数1<=, <=10
    int X[10][10]; // 一次元目:点の数, 二次元目:次元数

    cin >> N;
    cin >> D;

    for (int i = 0; i < N; i++) {
        for (int j = 0; j < D; j++) {
            cin >> X[i][j];
        }
    }

    int intDist = 0;
    int count = 0;
    for (int i = 0; i < N - 1; i++) {
        for (int j = i + 1; j < N; j++) {

            float sum = 0;
            for (int k = 0; k < D; k++) {
                sum += powf(abs(X[i][k] - X[j][k]), 2);
            }
            float dist = sqrt(sum);
            intDist = static_cast<int>(dist);
            if (static_cast<float>(intDist) == dist) {
                count++;
            }
        }
    }

    std::cout << count << std::endl;

    return 0;
}