1-D Prefix Sum

range: l r = prefix[r] - prefix[l-1]
       1 3 = 10 - 1 
           = 9

Algorithm:

        Read the Array Size n:
        Initialize an array arr of size n.
        Input the Elements into arr:

        For each i from 0 to n-1, input the value of arr[i].
        Construct the Prefix Sum Array pre:

        Initialize pre of size n.
        Set pre[0] = arr[0] (the first element).
        For each index i from 1 to n-1, calculate:
        pre[i] = pre[i-1] + arr[i]
        This step ensures that pre[i] holds the sum of all elements from 
        arr[0] to arr[i].
        Process Each Query:

        Read the integer q (number of queries).
        For each query:
        Input the range [l, r] (0-indexed).
        If l == 0, print pre[r] (sum from start to r).
        Else, print pre[r] - pre[l - 1] (sum from index l to r).

#include<bits/stdc++.h>

using namespace std;

int main()
{
    int n;
    cin >> n;

    int arr[n+1];

    // input array
    for(int i=0; i<n; i++)cin >> arr[i];

    // apply prefix
    int pre[n+1];
    pre[0] = arr[0];

    for(int i=1; i<n; i++)
    {
        pre[i] = pre[i+1] + arr[i];
    }

    // apply querie
   int q;
   cin >> q;

   while(q--)
   {
    int l, r;
    cin >> l >> r;

    if(l == 0)cout << pre[r] << endl;
    else cout << pre[r] - pre[l - 1] << endl;
   }
}