Utopian Tree

Problem Statement :

The Utopian Tree goes through 2 cycles of growth every year. Each spring, it doubles in height. Each summer, its height increases by 1 meter.

A Utopian Tree sapling with a height of 1 meter is planted at the onset of spring. How tall will the tree be after  growth cycles?

For example, if the number of growth cycles is n = 5, the calculations are as follows:

Period  Height
0          1
1          2
2          3
3          6
4          7
5          14

Function Description

Complete the utopianTree function in the editor below.

utopianTree has the following parameter(s):

int n: the number of growth cycles to simulate


int: the height of the tree after the given number of cycles

Input Format

The first line contains an integer, t, the number of test cases.
t subsequent lines each contain an integer, n, the number of cycles for that test case.

1 <= t <= 10
0 <= n <= 60

Solution :


                            Solution in C :

python 3  :

#!/usr/bin/env python

import sys

if __name__ == '__main__':
    T = int(sys.stdin.readline())
    for _ in range(T):
        N = int(sys.stdin.readline())
        height = 1
        for i in range(N):
            if i % 2 == 0:
                height *= 2
                height += 1

Java  :

import java.util.Scanner;
import java.util.Vector;

public class Solution {
    public static void main(String[] args) {
        Scanner stdin = new Scanner(System.in);
        int numCases = stdin.nextInt();
        Vector<Long> cache = new Vector<>();
        for (int i = 0; i < numCases; i++) {
            int numCycles = stdin.nextInt();
            while(cache.size() <= numCycles) {
                if (cache.size() % 2 == 0) {
                    cache.add(cache.lastElement() + 1);                
                } else {
                    cache.add(cache.lastElement() * 2);

C ++  :

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int main() {
    int T;
    cin >> T;
    for(int t = 0; t<T; t++){
        int N;
        cin >> N;
        int ht = 1;
        for(int i=0;i<N;i++){
            if(i%2==0) ht*=2;
            else ht++;
        cout << ht << endl;
    return 0;

C  :

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {

    /* Enter your code here. Read input from STDIN. Print output to STDOUT */ 
    int t,i,n;
       int h=1;
    return 0;

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