Let say,

f(n) = maximum calorie can be taken till nth day under the constraint

No, let's think of the case,

1. If Shivang takes food on ith day, he need to skip on (i-1)th day and sum with f(i-2)
2. Tale food on ith day, (i-1)th day and skip on (i-2)th day, sum up with f(i-3)
3. Skip ith day, so take f(i-1)

So,

f(i)=maximum(f(i-1),f(i-2)+arr[i],arr[i]+arr[i-1]+f(i-3) i>=3

For base case,
f(0)=arr[0]
f(1)=arr[0]+arr[1]
f(2)=maximum(arr[0]+arr[1],arr[1]+arr[2],arr[2]+arr[0])

Converting to Dynamic programming

Now, we need to convert the above recursion into dynamic programming to avoid the overlapping sub-problem re-computation

1)  Declare int DP[n];
2)  Fill with the base case
        DP[0]=arr[0];
        DP[1]=arr[0]+arr[1];
        DP[2]=std::max(arr[1]+arr[2],std::max(arr[0]+arr[2],DP[1]));
3)  Compute the other values
        for i=3 to n-1
            DP[i]=maximum(arr[i]+arr[i-1]+DP[i-3],arr[i]+DP[i-2],DP[i-1]);
        end for
4)  Return the result, DP[n-1];

C++ Implementation:

#include <bits/stdc++.h>
using namespace std;

int maximumCalorie(vector<int> arr, int n)
{

    int DP[n];
    // base case
    DP[0] = arr[0];
    DP[1] = arr[0] + arr[1];
    DP[2] = std::max(arr[1] + arr[2], std::max(arr[0] + arr[2], DP[1]));

    for (int i = 3; i < n; i++) {
        DP[i] = std::max(arr[i] + arr[i - 1] + DP[i - 3], std::max(arr[i] + DP[i - 2], DP[i - 1]));
    }

    return DP[n - 1];
}

int main()
{
    int t, n, item;
    
    cout << "Enter number of days\n";
    cin >> n;
    cout << "Enter calories\n";
    vector<int> a;
    
    for (int j = 0; j < n; j++) {
        scanf("%d", &item);
        a.push_back(item);
    }
    
    cout << "Maximum calorie sudhir can take: " << maximumCalorie(a, n) << endl;

    return 0;
}

Output:

Enter number of days:
8 
Enter calories:
3 2 3 2 3 5 1 3 
Maximum calorie sudhir can take: 17