How to solve the "Utopian Tree" - c++

Question

The Utopian Tree goes through 2 cycles of growth every year. The first growth cycle occurs during the spring, when it doubles in height. The second growth cycle occurs during the summer, when its height increases by 1 meter.
Now, a new Utopian Tree sapling is planted at the onset of spring. Its height is 1 meter. Can you find the height of the tree after N growth cycles?
#include <iostream>
using namespace std;
int height(int n) {
int h[61],i;
h[0]=1;
for(i=1;i<61;i++)
{ if (i%2!=0)
h[i]=h[i-1]*2;
else h[i]=h[i-1]+1;
}
cout<<h[n];
return 0;
}
int main() {
int T;
cin >> T;
while (T--) {
int n;
cin >> n;
cout << height(n) << endl;
}
}

Answer 1

Here is a simple solution that avoids too much calculations.
Notice the following:
n: Height: Hint:
0 1 2^0
1 2*1=2 2^1
2 2+1=3 2^2 -1
3 2*3=6 2^3 -2
4 6+1=7 2^3 -1
5 2*7=14 2^4 -2
6 14+1=15 2^4 -1
7 2*15=30 2^5 -2
8 30+1=31 2^5 -1
9 2*31=62 2^6 -2
10 62+1=63 2^6 -1
11 2*63=126 2^7 -2
12 126+1=127 2^7 -1
13 2*127=254 2^8 -2
14 254+1=255 2^8 -1
15 2*255=510 2^9 -2
16 510+1=511 2^9 -1
and so on and so forth...
That means that we can use bit shift and avoid for loops etc. Here is a simple solution:
int main(){
//number of test cases
int t;
cin >> t;
for(int i= 0; i< t; ++i){
//number of cycles for the tree growth
int n;
cin >> n;
if (n == 0)
cout << 1 << endl;
else if (n == 1)
cout << 2 << endl;
else if (n > 1){
cout << ((1 << ( (n & 1) ? ((n+1)/2) : (n/2)) + 1) - ((n & 1) ? 2 : 1)) << endl;
}
}
return 0;
}
The key is to categorize number of cycles in odd and even and process accordingly.

Answer 2

You should return h[n] instead of 0 here:
int height(int n) {
int h[61],i;
h[0]=1;
for(i=1;i<61;i++)
{ if (i%2!=0)
h[i]=h[i-1]*2;
else h[i]=h[i-1]+1;
}
//cout<<h[n]<<endl;
return h[n];
}
Also, note that it was printing 10 and 20 because the value of h[n] was actually 1 and 2 itself but the function was returning 0 itself.
Due to the statement:
cout << h[n];
It was printing 1(and 2 for next call) and then due to statement:
cout<<height(n);
it was printing the 0(and 0 for next call). Thus, it was printing 1 and 0 for height[0] and then printing 2 and 0 for height[0] .
Summary:
Return h[n] from function instead of 0.
Also, use tab or newline with cout to minimize the confusion.
Why to print the value in both height() and main(). Simply do it in main()

Answer 3

The code should produce 10 and 20 based on your program if you enter 2 0 1. You might want to have a look at what you output and what you return from your function plus what you do with the result.
Note that you should also always check that your inputs were actually successful e.g. using if (std::cin >> T) { ... }. However, that's not the problem in your code.

Answer 4

The function int height(int ) should return h[n]; instead of printing h[n] (cout << h[n];) and then return 0 (which adds 0 next to your desired output of 1,2). Using your code, you should modify it as follows in order to make it work:
#include <iostream>
using namespace std;
int height(int n) {
int h[61],i;
h[0]=1;
for(i=1;i<61;i++)
{ if (i%2!=0)
h[i]=h[i-1]*2;
else h[i]=h[i-1]+1;
}
//cout<<h[n];
return h[n];
}
int main() {
int T;
cin >> T;
while (T--) {
int n;
cin >> n;
cout << height(n) << endl;
}
}

Answer 5

How about array of functions?
typedef int(*FunctionPrt)(int);
int utopianTree(int n) {
int initialHeight = 1;
FunctionPrt functions[] = {
[](int initialHeight) { return initialHeight * 2; },
[](int initialHeight) { return initialHeight + 1; }
};
for (int cycles = 0, i = 0; cycles < n; i = (i + 1) % 2, ++cycles) {
initialHeight = functions[i](initialHeight);
}
return initialHeight;
}