Solução do Problema Polygon Area do CSES

Problem

Your task is to calculate the area of a given polygon. The polygon consists of n vertices \((x_1,y_1),(x_2,y_2),\dots,(x_n,y_n)\). The vertices \((x_i,y_i)\) and \((x_{i+1},y_{i+1})\) are adjacent for \(i=1,2,\dots,n-1\), and the vertices \((x_1,y_1)\) and \((x_n,y_n)\) are also adjacent.

Input

The first input line has an integer n: the number of vertices. After this, there are n lines that describe the vertices. The ith such line has two integers \(x_i\) and \(y_i\). You may assume that the polygon is simple, i.e., it does not intersect itself.

Output

Print one integer: \(2a\) where the area of the polygon is a (this ensures that the result is an integer).

#include <bits/stdc++.h>
using namespace std;
 
#define int long long
 
 
struct Ponto
{
    int x;
    int y;
};
  
void solve()
{
    int n;
    cin >> n;
    vector <Ponto> poligono;
    for(int i = 0 ;  i< n; i++)
    {
        int x, y;
        cin >> x >> y;
        Ponto p = {x,y};
        poligono.push_back(p);
    }
 
 
    int area = 0.0;
 
    for(int i = 0; i < n; i++)
    {
        Ponto p1 = poligono[i];
        Ponto p2 = poligono[(i + 1) % n];
        area += (p1.x * p2.y) - (p1.y * p2.x);
    }
 
    if(area < 0)
        cout << area * -1 << endl;
    else
        cout << area << endl;  
}
 
signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
 
    int t = 1;
    while (t--)
    {
        solve();
    }
 
    return 0;
}