Need to design a c++ program do the following. Problem is listed and is followed

Question

Need to design a c++ program do the following. Problem is listed and is followed by intended output.

need to solve recurrence t(n)=t(n-1)+n, for n>1, t(1)=a; where a is some constant of type double Enter a and n

1. Find t(a, n) using recursion What is complexity of your solution, when you use recursion?

2. find t(a,n)==f(a,n) in O(1) complexity Provide code for both solutions. for n<1, t(a,n) is not defined, so throw exception if user enters not number, and throw exception if user enters number less than 1 catch both exceptions and ask user to re-enter n value

************************************************************************

intended output:

t(1)=a Enter constant double a = w

you entered not double type value

t(1)=a Enter constant double a = r

t(1)=a Enter constant double a = 0.5

Enter int n =0

you entered 0 number is less than 1

t(1)=a Enter constant double a = 4

Enter int n =-3

you entered -3 number is less than 1

t(1)=a Enter constant double a = 0.5

Enter int n =8

recursive       t(0.5,8)=35.5

complexity O(1) f(0.5,8)=35.5

Do you want to repeate? y/n: y

t(1)=a Enter constant double a = r

t(1)=a Enter constant double a = 1

Enter int n =5

recursive       t(1,5)=15

complexity O(1) f(1,5)=15

Do you want to repeat? y/n: n

Explanation / Answer

// C++ program to find the element

// occurring odd number of times

#include<bits/stdc++.h>

using namespace std;

// Funtion to find the element

// occurring odd number of times

int getOddOccurrence(int arr[], int arr_size)

{

    for (int i = 0; i < arr_size; i++) {

        int count = 0;

        for (int j = 0; j < arr_size; j++)

        {

            if (arr[i] == arr[j])

                count++;

        }

        if (count % 2 != 0)

            return arr[i];

    }

    return -1;

}

// driver code

int main()

    {

        int arr[] = { 2, 3, 5, 4, 5, 2,

                      4, 3, 5, 2, 4, 4, 2 };

        int n = sizeof(arr) / sizeof(arr[0]);

        // Function calling

        cout << getOddOccurrence(arr, n);

        return 0;

    }

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