Dynamic programming to solve the fibwords problem - recursion

Question

Problem Statement: The Fibonacci word sequence of bit strings is defined as:
F(0) = 0, F(1) = 1
F(n − 1) + F(n − 2) if n ≥ 2
For example : F(2) = F(1) + F(0) = 10, F(3) = F(2) + F(1) = 101, etc.
Given a bit pattern p and a number n, how often does p occur in F(n)?
Input:
The first line of each test case contains the integer n (0 ≤ n ≤ 100). The second line contains the bit
pattern p. The pattern p is nonempty and has a length of at most 100 000 characters.
Output:
For each test case, display its case number followed by the number of occurrences of the bit pattern p in
F(n). Occurrences may overlap. The number of occurrences will be less than 2^63.
Sample input: 6 10 Sample output: Case 1: 5
I implemented a divide and conquer algorithm to solve this problem, based on the hints that I found on the internet: We can think of the process of going from F(n-1) to F(n) as a string replacement rule: every '1' becomes '10' and '0' becomes '1'. Here is my code:
#include <string>
#include <iostream>
using namespace std;
#define LL long long int
LL count = 0;
string F[40];
void find(LL n, char ch1,char ch2 ){//Find occurences of eiher "11" / "01" / "10" in F[n]
LL n1 = F[n].length();
for (int i = 0;i+1 <n1;++i){
if (F[n].at(i)==ch1&&F[n].at(i+1)==ch2) ++ count;
}
}
void find(char ch, LL n){
LL n1 = F[n].length();
for (int i = 0;i<n1;++i){
if (F[n].at(i)==ch) ++count;
}
}
void solve(string p, LL n){//Recursion
// cout << p << endl;
LL n1 = p.length();
if (n<=1&&n1>=2) return;//return if string pattern p's size is larger than F(n)
//When p's size is reduced to 2 or 1, it's small enough now that we can search for p directly in F(n)
if (n1<=2){
if (n1 == 2){
if (p=="00") return;//Return since there can't be two subsequent '0' in F(n) for any n
else find(n,p.at(0),p.at(1));
return;
}
if (n1 == 1){
if (p=="1") find('1',n);
else find('0',n);
return;
}
}
string p1, p2;//if the last character in p is 1, we can replace it with either '1' or '0'
//p1 stores the substring ending in '1' and p2 stores the substring ending in '0'
for (LL i = 0;i<n1;++i){//We replace every "10" with 1, "1" with 0.
if (p[i]=='1'){
if (p[i+1]=='0'&&(i+1)!= n1){
if (p[i+2]=='0'&&(i+2)!= n1) return;//Return if there are two subsequent '0'
p1.append("1");//Replace "10" with "1"
++i;
}
else {
p1.append("0");//Replace "1" with "0"
}
}
else {
if (p[i+1]=='0'&&(i+1)!= n1){//Return if there are two subsequent '0'
return;
}
p1.append("1");
}
}
solve(p1,n-1);
if (p[n1-1]=='1'){
p2 = p1;
p2.back() = '1';
solve(p2,n-1);
}
}
main(){
F[0] = "0";F[1] = "1";
for (int i = 2;i<38;++i){
F[i].append(F[i-1]);
F[i].append(F[i-2]);
}//precalculate F(0) to F(37)
LL t = 0;//NumofTestcases
int n; string p;
while (cin >> n >> p) {
count = 0;
solve(p,n);
cout << "Case " << ++t << ": " << count << endl;
}
}
The above program works fine, but with small inputs only. When i submitted the above program to codeforces i got an answer wrong because although i shortened the pattern string p and reduces n to n', the size of F[n'] is still very large (n'>=50). How can i modify my code to make it works in this case, or is there another approach (such as dynamic programming?). Many thanks for any advice.
More details about the problem can be found here: https://codeforces.com/group/Ir5CI6f3FD/contest/273369/problem/B

Answer 1

I don't have time now to try to code this up myself, but I have a suggested approach.
First, I should note, that while that hint you used is certainly accurate, I don't see any straightforward way to solve the problem. Perhaps the correct follow-up to that would be simpler than what I'm suggesting.
My approach:
Find the first two ns such that length(F(n)) >= length(pattern). Calculating these is a simple recursion. The important insight is that every subsequent value will start with one of these two values, and will also end with one of them. (This is true for all adjacent values -- for any m > n, F(m) will begin either with F(n) or with F(n - 1). It's not hard to see why.)
Calculate and cache the number of occurrences of the pattern in this these two Fs, but whatever index shifting technique makes sense.
For F(n+1) (and all subsequent values) calculate by adding together
The count for F(n)
The count for F(n - 1)
The count for those spanning both F(n) and F(n - 1). We can achieve that by testing every breakdown of pattern into (nonempty) prefix and suffix values (i.e., splitting at every internal index) and counting those where F(n) ends in prefix and F(n - 1) starts with suffix. But we don't have to have all of F(n) and F(n - 1) to do this. We just need the tail of F(n) and the head of F(n - 1) of the length of the pattern. So we don't need to calculate all of F(n). We just need to know which of those two initial values our current one ends with. But the start is always the predecessor, and the end oscillates between the previous two. It should be easy to keep track.
The time complexity then should be proportional to the product of n and the length of the pattern.
If I find time tomorrow, I'll see if I can code this up. But it won't be in C -- those years were short and long gone.
Collecting the list of prefix/suffix pairs can be done once ahead of time