The solution requires the knowledge of basic geometry. The question is asking for the Euclidean distance between two points in ~N~-dimensional space.

First of all, it should be known that the formula for calculating Euclidean distance in ~2~-dimensional space is:

$$\displaystyle d\ =\ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$

For the first subtask, the ~2~-dimensional distance formula can be expanded to ~3~ dimensions.

$$\displaystyle d\ =\ \sqrt{(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2})^2 + (z_2 - z_1)^2}$$

$$\displaystyle d\ =\ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$$

For the second and last subtasks, the ~2~-dimensional distance formula can be expanded to ~N~-dimensions.

The formula for calculating Euclidean distance in ~N~-dimensional space is

$$\displaystyle d = \displaystyle \sqrt{\sum_{i=1}^N (b_i - a_i)^2}$$

The proof is left as an exercise for the reader.

The second subtask was specifically dedicated to those who used 32-bit floating-point variables instead of 64-bit floating-point variables in their calculations.

Time Complexity: ~O(N)~

Sample Solution - D

import std.math;
import std.stdio;

void main()
{
    int N;
    double distance = 0;
    double[1000] a, b;
    scanf("%d", &N);
    foreach(x;0..N)
        scanf("%lf", &a[x]);
    foreach(x;0..N)
        scanf("%lf", &b[x]);
    foreach(x;0..N)
        distance += (b[x]-a[x])*(b[x]-a[x]);
    printf("%lf\n", sqrt(distance));
}