Envío 618

Código

// https://codeo.app/problemas/0xf2-partir-un-arreglo-en-2
// Calculate index to which right part sum is negative and left side positive (bigger left)

#include <iostream>
#include <vector>

using namespace std;

int main()
{
    int n, partitionIndex, rightSum;
    vector<int> arr, leftSumAt;

    // Read inputs
    cin >> n;
    arr = vector<int>(n);
    leftSumAt = vector<int>(n);
    int sumCache = 0;

    for (int i = 0; i < n; i++)
    {
        cin >> arr[i];
        sumCache += arr[i];
        leftSumAt[i] = sumCache;
    }

    partitionIndex = -1;
    rightSum = 0;
    // Navigate from right to left, checking left when right part sum is neg
    for (int i = n - 1; i > 0; i--)
    {
        rightSum += arr[i];
        if (rightSum < 0 && leftSumAt[i - 1] > 0)
        {
            partitionIndex = i;
        }
    }

    // Output
    if (partitionIndex == -1)
    {
        cout << "Impossible" << endl;
    }
    else
    {
        cout << partitionIndex << endl;
    }

    return 0;
}