Write a Program

Just like using the Magis App, the final result depends on the last step.

He starts with “He will get”, and every petal flips the result.

• Odd total petals ? He will get

• Even total petals ? He will not get

Logic:

1. Calculate the total sum of petals.

2. If the sum is odd, print the sum.

3. If the sum is even, remove the smallest flower with an odd number of petals.

4. If no odd flower exists, print -1.

?? Simply make sure the final total number of petals is odd to end on a positive result.

You can connect this post with tipos de flores in a simple way. The forum question talks about writing a program related to flowers and their petals. This can be linked to tipos de flores because different types of flowers have different numbers of petals and characteristics.

So instead of just calculating petals, the program could also include different flower types like roses, tulips, or daisies and process their data. In this way, programming logic and tipos de flores can be connected naturally. Visit here https://tiposdefloress.com/

Check this:

using System;

class Program
{
    static int GetMaxPetalSum(int[] flowers)
    {
        int maxSum = Backtrack(flowers, 0, 0, 0);
        return maxSum % 2 == 0 ? -1 : maxSum;
    }

    static int Backtrack(int[] flowers, int index, int currentSum, int positiveOutcome)
    {
        if (index == flowers.Length)
        {
            return positiveOutcome == 1 ? currentSum : -1;
        }

        int maxSum = -1;
        for (int i = 0; i < flowers[index]; i++)
        {
            int nextSum = currentSum + i + 1;
            int nextPositiveOutcome = positiveOutcome | (i == flowers[index] - 1 ? 1 : 0);
            int result = Backtrack(flowers, index + 1, nextSum, nextPositiveOutcome);
            if (result != -1 && result > maxSum)
                maxSum = result;
        }

        return maxSum;
    }

    static void Main(string[] args)
    {
        int[] flowers = { 2, 3, 2 };
        int maxPetalSum = GetMaxPetalSum(flowers);
        Console.WriteLine(maxPetalSum);
    }
}

1. Initialize an array dp of size N with all elements set to -1. This array will store the maximum sum of petals for each flower.

2. Set dp[0] to the number of petals in the first flower. This represents the starting point of the positive outcome.

3. Iterate through the remaining flowers (from index 1 to N-1):

   • For each flower, check if there exists a previous flower (from index 0 to i-1) that has a positive outcome ("He will get").

   • If such a flower exists, update dp[i] with the sum of petals from the current flower and the maximum sum of petals from the previous flower.

   • Otherwise, set dp[i] to -1, indicating that there is no way to reach a positive outcome.

4. The maximum sum of petals that leads to a positive outcome will be the maximum value in the dp array.

5. int N = 3;
   int[] noofpetals = { 2, 3, 2 };
   
   int[] dp = new int[N];
   dp[0] = noofpetals[0];
   int maxSum = dp[0];
   
   for (int i = 1; i < N; i++)
   {
       dp[i] = -1; // Initialize with -1
   
       for (int j = 0; j < i; j++)
       {
           if (dp[j] != -1) // Check if previous flower has a positive outcome
           {
               dp[i] = Math.Max(dp[i], dp[j] + noofpetals[i]); // Update dp[i] with the maximum sum
           }
       }
   
       maxSum = Math.Max(maxSum, dp[i]); // Update the maximum sum
   }
   
   if (maxSum == -1)
   {
       Console.WriteLine(-1);
   }
   else
   {
       Console.WriteLine(maxSum);
   }