how to know all the length of cuts of rod in rod cutting algorithm ? (dynamic programming)

i know the rod cutting algorithm. The c++ implementation is below:

// A Dynamic Programming solution for Rod cutting problem
#include<stdio.h>
#include<limits.h>

// A utility function to get the maximum of two integers
int max(int a, int b) { return (a > b)? a : b;}

/* Returns the best obtainable price for a rod of length n and
   price[] as prices of different pieces */
int cutRod(int price[], int n)
{
   int val[n+1];
   val[0] = 0;
   int i, j;

   // Build the table val[] in bottom up manner and return the last entry
   // from the table
   for (i = 1; i<=n; i++)
   {
       int max_val = INT_MIN;
       for (j = 0; j < i; j++)
         max_val = max(max_val, price[j] + val[i-j-1]);
       val[i] = max_val;
   }

   return val[n];
}

/* Driver program to test above functions */
int main()
{
    int arr[] = {1, 5, 8, 9, 10, 17, 17, 20};
    int size = sizeof(arr)/sizeof(arr[0]);
    printf("Maximum Obtainable Value is %d\n", cutRod(arr, size));
    getchar();
    return 0;
}

the output is:

Maximum Obtainable Value is 22

My question is i can find the maximum value (price) of cutting a particular length rod but how can i find the length of cuts in that particular rod ?

You may update the function cutRod so that while going up in your bottom-up approach, you try also to remember where you got the optimal result from. (i.e. the last rod you cut to reach that step) Once you're done going up, before returning, you may start from the final point you arrived at and trace back each rod you cut until you arrive at length 0, which was also the base for your bottom-up approach.

You may find a crude implementation below.

int numRodsUsed;
int cutRod(int price[], int rods[], int n)
{
    int val[n+1];
    int lastRod[n+1];
    val[0] = 0;
    int i, j;

    // Build the table val[] in bottom up manner and return the last entry
    // from the table
    for (i = 1; i<=n; i++)
    {
        int max_val = INT_MIN;
        int best_rod_len = -1;
        for (j = 0; j < i; j++)
        {
            if(max_val < price[j] + val[i-j-1])
            {
                max_val = price[j] + val[i-j-1];
                best_rod_len = j;
            }
        }
        val[i] = max_val;
        lastRod[i] = best_rod_len + 1;
    }

    for (i = n, j = 0; i>0; i -= lastRod[i])
    {
        rods[j++] = lastRod[i];
    }
    numRodsUsed = j;

    return val[n];
}

int main()
{
    int arr[] = {1, 5, 8, 9, 10, 17, 17, 20};
    int size = sizeof(arr)/sizeof(arr[0]);
    int rods[size+1];
    int i;
    printf("Maximum Obtainable Value is %d\n", cutRod(arr, rods, size));
    printf("Rod lengths are: %d", rods[0]);
    for(i = 1; i < numRodsUsed; i++)
    {
        printf(" %d", rods[i]);
    }
    printf("\n");
    getchar();
    return 0;
}