Frogger crossing the river [recursive method]

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topic description

There is a river with a stone pier on the left ( $Area\; A$ ) has numbers$1，2，3，4，\dots ,n$ of$n$ frogs, there are$k$ lotus leaves ($Area\; C$ ), and$h$ stone piers ($Area\; D$ ), with a stone pier on the right ($Area\; B$ ), as shown in the figure below. $n$ frogs want to cross the river (from the stone pier$A$ to the stone pier on the right bank$B$ ), the rule is:

• The stone pier can bear any number of frogs, and the lotus leaf can only bear one frog (regardless of size);
• Frogs can: $A$→$B$ (available from$A$ jumps to$B$ , same below),$A$→$C$，$A$→$D$，$C$→$B$，$D$→$B$，$D$→$C$，$C$→$D$；
• When there are multiple frogs on a stone pier, the frog on it can only jump to the frog that is 1 size larger than it.

Your task is to give $h$，$k$ , calculate and output how many frogs can cross the river smoothly according to the above rules?

Input and output samples

input sample

$2$ $3$

There are 2 2 in the middle of the river$2$ stone grate,$3$ lotus leaves

output sample

$16$

Up to $16$ frogs can cross the river according to the rules

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recursive solution

the key here
From specific to general, the derivation process is as follows:
$f(0,0)=1$
$f(0,k)=k+1;$ 　　　　　 (e.g.$k=3$ when there are$4$ frogs can cross the river)
$f(1,k)=2(k+1);$ 　　　　　 (recursion thinking)
$f(2,k)=2(k+1)\times 2=2^2\times (k+1);$
...
and so on to get the recursive formula:$F(i,k)=F(i-1,k)\times 2=2^i\times (k+1);$
...
therefore,$f(h,k)=2^h\times (k+1)$

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AC code

#include <bits/stdc++.h>
using namespace std;
long long h,k,ans=1;
int main()
{
    
    
	cin >>h >>k;
	while (h--)
		ans*=2;
	ans*=(k+1);
	cout <<ans;
	return 0;
}