C++ Program to check we can remove all stones by selecting boxes

Suppose we have an array A with N elements. Consider there are N boxes and they are arranged in a circle. The ith box contains A[i] stones. We have to check whether we can remove all stones from the boxes by repeatedly performing the operation: Select a box say ith box. For each j in range 1 to N, remove exactly j stones from the (i+j)th box. Here (N+k)th box is termed as kth box. This operation cannot be performed if a box does not contain sufficient number of stones.

So, if the input is like A = [4, 5, 1, 2, 3], then the output will be True, because we can remove all stones by starting from second box.

To solve this, we will follow these steps −

n := size of A
Define an array a of size (n + 1)
Define an array b of size (n + 1)
sum := 0, p := n * (n + 1)
for initialize i := 1, when i 
Example
Let us see the following implementation to get better understanding −
#include 
using namespace std;
bool solve(vector A) {
   int n = A.size();
   vector a(n + 1);
   vector b(n + 1);
   int sum = 0, p = n * (n + 1) / 2;
   for (int i = 1; i  A = { 4, 5, 1, 2, 3 };
   cout 

Input
{ 4, 5, 1, 2, 3 }

Output
1