B - Traveling Takahashi Problem (AtCoder Beginner Contest 375)

Problem Link: B - Traveling Takahashi Problem (AtCoder Beginner Contest 375)

Approach

The code takes a list of points and calculates the total distance between them. It first reads the points, adds the origin $(0, 0)$ at the start, reverses the list, adds origin at end again, updates the size of $n$ and and then computes the distance between each pair of consecutive points. Finally, it prints the total distance with precision $20$ .

Code Implementation

#include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;

signed main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);

    int t=1;
    // cin >> t;

    while (t--) {
        int n;
        cin >> n;
        vector<pair<int, int>> v(n);
        for (auto &x:v){
            cin >> x.first >> x.second;
        }
        v.emplace_back(0,0);
        reverse(v.begin(),v.end());
        v.emplace_back(0,0);
        n=size(v);
        long double ans=0;
        for (int i=0; i+1<n; i++) {
            const int x=v[i].first-v[i+1].first, y=v[i].second-v[i+1].second;
            ans+=sqrt(x*x+y*y);
        }
        cout << fixed << setprecision(20) << ans << endl;
    }
    return 0;

}

import math

def solve():
    n = int(input())
    points = []
    for _ in range(n):
        points.append(list(map(int, input().split())))

    # Add origin at start and end
    # Note: The C++ solution logic effectively adds origin at both ends
    # (indirectly via reverse logic), here is the direct equivalent:
    path = [[0, 0]] + points + [[0, 0]]

    ans = 0.0
    for i in range(len(path) - 1):
        x1, y1 = path[i]
        x2, y2 = path[i+1]
        dx = x1 - x2
        dy = y1 - y2
        ans += math.sqrt(dx*dx + dy*dy)

    print(f"{ans:.20f}")

if __name__ == "__main__":
    solve()