Hello reader!↵
↵
I came across this problem and I'm not sure how to approach it. Any hints would be appreciated!↵
↵
**Problem:**↵
↵
The Fibonacci sequence is defined as F(0) = 1, F(1) = 1, F(n) = F(n-1) + F(n-2).↵
So: 1, 1, 2, 3, 5, 8, 13, 21, 34, ...↵
↵
Given N, count the number of ways to write N as a product of Fibonacci numbers, where each factor is > 1. Order does not matter (2×4 and 4×2 are the same representation).↵
↵
**Example:**↵
Input:↵
4↵
2↵
7↵
40↵
64↵
Output:↵
1↵
0↵
2↵
3↵
↵
**Constraints:**↵
- T ≤ 50 test cases↵
- N can be up to 10^18↵
↵
I have no idea where to start. How do you approach this kind of problem?↵
↵
Thank you!↵
↵
update : the problem was solved↵
↵
**solotion 1: Top-Down DP with Memoization** author [user:MEDAAA,2026-05-25]↵
↵
<spoiler summary="Idea">↵
↵
Instead of finding all divisors of N (which takes $O(\sqrt{N})$ time and causes TLE for $N=10^{18}$), we recursively divide N by Fibonacci numbers.↵
↵
To avoid counting permutations multiple times, we enforce an ordering constraint: we only use Fibonacci numbers from a certain index onwards, ensuring each factorization is counted exactly once.↵
↵
**Algorithm:**↵
1. Precompute all Fibonacci numbers ≤ $10^{18}$↵
2. Use recursive DP with memoization: `dp(N, idx)` = number of ways to factorize N using Fibonacci numbers from index `idx` onwards↵
3. For each Fibonacci number at index `idx`:↵
— Option 1: Skip it and move to `idx + 1`↵
— Option 2: If it divides N, use it and recurse on `N / fib[idx]` (same index to allow repetition)↵
↵
**Time Complexity:** $O(T \cdot \text{Fib\_count}^2 \cdot \log N)$ due to memoization ↵
**Space Complexity:** $O(\text{Fib\_count} \cdot \log N)$ for memoization table↵
↵
</spoiler>↵
↵
<spoiler summary="Code">↵
↵
```cpp↵
#include <iostream>↵
#include <vector>↵
#include <map>↵
↵
using namespace std;↵
↵
vector<long long> fibs;↵
map<pair<long long, int>, long long> memo;↵
↵
void precompute() {↵
long long a = 1, b = 2;↵
// Generate Fibonacci numbers > 1 up to 10^18↵
// Sequence starts: 2, 3, 5, 8, 13...↵
fibs.push_back(b);↵
while (true) {↵
long long next_fib = a + b;↵
if (next_fib > 1e18) break;↵
fibs.push_back(next_fib);↵
a = b;↵
b = next_fib;↵
}↵
}↵
↵
// Top-Down DP: count ways to factorize N using fibs[idx..end]↵
long long dp(long long N, int idx) {↵
// Base case: N reduced to 1 = found 1 valid factorization↵
if (N == 1) return 1;↵
↵
// No Fibonacci numbers left but N > 1 = invalid path↵
if (idx >= fibs.size()) return 0;↵
↵
// Check memoization↵
if (memo.count({N, idx})) return memo[{N, idx}];↵
↵
long long ways = 0;↵
↵
// Option 1: Don't use fibs[idx], move to next Fibonacci↵
ways += dp(N, idx + 1);↵
↵
// Option 2: Use fibs[idx] if it divides N↵
// Keep same idx to allow using this Fibonacci multiple times↵
if (N % fibs[idx] == 0) {↵
ways += dp(N / fibs[idx], idx);↵
}↵
↵
return memo[{N, idx}] = ways;↵
}↵
↵
void solve() {↵
long long N;↵
cin >> N;↵
cout << dp(N, 0) << "\n";↵
}↵
↵
int main() {↵
ios_base::sync_with_stdio(false);↵
cin.tie(NULL);↵
↵
precompute();↵
↵
int t;↵
if (cin >> t) {↵
while (t--) solve();↵
}↵
return 0;↵
}↵
```↵
↵
</spoiler>↵
↵
**solution 2: Greedy + Math** author:[user:phsads,2026-05-25]↵
↵
<spoiler summary="Idea">↵
↵
By **Carmichael's theorem**, large Fibonacci numbers (except 2, 3, 8, 144) contain unique prime factors not found in smaller Fibonacci numbers. ↵
↵
This means:↵
- If N is divisible by a large Fibonacci number, we **must** use it in the factorization↵
- We greedily divide N by large Fibonacci numbers first↵
- After this greedy phase, N becomes a product of powers of 2 and 3 only↵
- We count valid combinations of {2, 3, 8, 144}↵
↵
**Algorithm:**↵
1. Generate Fibonacci numbers, excluding 2, 3, 8, 144 (the "exception" numbers)↵
2. Greedily divide N by large Fibonacci numbers (from largest to smallest)↵
3. Factorize remaining N into $2^p \times 3^q$↵
4. Count ways to form $2^p \times 3^q$ using {2, 3, 8, 144}:↵
— Iterate over count of 144 used (each 144 = $2^4 \times 3^2$)↵
— Remaining power of 2 must be formed using {2, 8}↵
— Number of ways to form remaining power P of 2: $\lfloor P/3 \rfloor + 1$↵
↵
**Time Complexity:** Nearly $O(\log N)$ per query ↵
**Space Complexity:** $O(\text{Fib\_count})$↵
↵
</spoiler>↵
↵
<spoiler summary="Code">↵
↵
```cpp↵
#include <iostream>↵
#include <vector>↵
#include <algorithm>↵
↵
using namespace std;↵
↵
vector<long long> greedy_fibs;↵
↵
void precompute() {↵
long long f[90];↵
f[1] = 1; f[2] = 1;↵
↵
// Generate Fibonacci sequence up to 10^18↵
for (int i = 3; i <= 87; i++) {↵
f[i] = f[i-1] + f[i-2];↵
↵
// Exclude exceptions: F_3=2, F_4=3, F_6=8, F_12=144↵
// These don't have unique prime factors↵
if (i != 3 && i != 4 && i != 6 && i != 12) {↵
greedy_fibs.push_back(f[i]);↵
}↵
}↵
↵
// Sort in descending order for greedy approach↵
sort(greedy_fibs.rbegin(), greedy_fibs.rend());↵
}↵
↵
void solve() {↵
long long N;↵
cin >> N;↵
↵
// Step 1: Greedily divide N by large Fibonacci numbers↵
// (they have unique prime factors, so MUST be used if they divide N)↵
for (long long fib : greedy_fibs) {↵
while (N % fib == 0) {↵
N /= fib;↵
}↵
}↵
↵
// Step 2: Factorize remaining N into 2^p * 3^q↵
long long p = 0, q = 0;↵
while (N % 2 == 0) { p++; N /= 2; }↵
while (N % 3 == 0) { q++; N /= 3; }↵
↵
// If remainder is not 1, it has other prime factors → impossible↵
if (N != 1) {↵
cout << 0 << "\n";↵
return;↵
}↵
↵
// Step 3: Count combinations using {2, 3, 8, 144}↵
// where: 144 = 2^4 * 3^2, 8 = 2^3, 3 = 3^1, 2 = 2^1↵
long long ans = 0;↵
↵
// Try all possible counts of 144↵
for (long long c144 = 0; c144 * 4 <= p && c144 * 2 <= q; c144++) {↵
↵
// Remaining power of 2 after using c144 copies of 144↵
long long P_remaining = p - c144 * 4;↵
↵
// Ways to form P_remaining using 8 (2^3) and 2 (2^1)↵
// This is: how many ways to write P as sum of 3's and 1's↵
ans += (P_remaining / 3) + 1;↵
}↵
↵
cout << ans << "\n";↵
}↵
↵
int main() {↵
ios_base::sync_with_stdio(false);↵
cin.tie(NULL);↵
↵
precompute();↵
↵
int t;↵
if (cin >> t) {↵
while (t--) solve();↵
}↵
return 0;↵
}↵
```↵
↵
</spoiler>↵
↵
---↵
↵
## Complexity Comparison↵
↵
| Approach | Time per Query | Space | Notes |↵
|----------|---|---|---|↵
| DP Memoization | $O(\text{Fib}^2 \log N)$ | $O(\text{Fib} \log N)$ | More intuitive, general DP |↵
| Greedy + Math | $O(\log N)$ | $O(\text{Fib})$ | Faster, uses number theory |↵
↵
## Credits↵
↵
Thanks to [user:MEDAAA,2026-05-25], [user:phsads,2026-05-25], [user:Ahmed-Nawaz,2026-05-25], [user:InfiniteLoops1730,2026-05-25], [user:VladiG,2026-05-25], [user:Metall1cA,2026-05-25], [user:helpT27,2026-05-25], and others for all the discussions and ideas in the comments. Really appreciate the different approaches and insights — from DP attempts to number theory explanations.
↵
I came across this problem and I'm not sure how to approach it. Any hints would be appreciated!↵
↵
**Problem:**↵
↵
The Fibonacci sequence is defined as F(0) = 1, F(1) = 1, F(n) = F(n-1) + F(n-2).↵
So: 1, 1, 2, 3, 5, 8, 13, 21, 34, ...↵
↵
Given N, count the number of ways to write N as a product of Fibonacci numbers, where each factor is > 1. Order does not matter (2×4 and 4×2 are the same representation).↵
↵
**Example:**↵
Input:↵
4↵
2↵
7↵
40↵
64↵
Output:↵
1↵
0↵
2↵
3↵
↵
**Constraints:**↵
- T ≤ 50 test cases↵
- N can be up to 10^18↵
↵
I have no idea where to start. How do you approach this kind of problem?↵
↵
Thank you!↵
↵
update : the problem was solved↵
↵
**solotion 1: Top-Down DP with Memoization** author [user:MEDAAA,2026-05-25]↵
↵
<spoiler summary="Idea">↵
↵
Instead of finding all divisors of N (which takes $O(\sqrt{N})$ time and causes TLE for $N=10^{18}$), we recursively divide N by Fibonacci numbers.↵
↵
To avoid counting permutations multiple times, we enforce an ordering constraint: we only use Fibonacci numbers from a certain index onwards, ensuring each factorization is counted exactly once.↵
↵
**Algorithm:**↵
1. Precompute all Fibonacci numbers ≤ $10^{18}$↵
2. Use recursive DP with memoization: `dp(N, idx)` = number of ways to factorize N using Fibonacci numbers from index `idx` onwards↵
3. For each Fibonacci number at index `idx`:↵
— Option 1: Skip it and move to `idx + 1`↵
— Option 2: If it divides N, use it and recurse on `N / fib[idx]` (same index to allow repetition)↵
↵
**Time Complexity:** $O(T \cdot \text{Fib\_count}^2 \cdot \log N)$ due to memoization ↵
**Space Complexity:** $O(\text{Fib\_count} \cdot \log N)$ for memoization table↵
↵
</spoiler>↵
↵
<spoiler summary="Code">↵
↵
```cpp↵
#include <iostream>↵
#include <vector>↵
#include <map>↵
↵
using namespace std;↵
↵
vector<long long> fibs;↵
map<pair<long long, int>, long long> memo;↵
↵
void precompute() {↵
long long a = 1, b = 2;↵
// Generate Fibonacci numbers > 1 up to 10^18↵
// Sequence starts: 2, 3, 5, 8, 13...↵
fibs.push_back(b);↵
while (true) {↵
long long next_fib = a + b;↵
if (next_fib > 1e18) break;↵
fibs.push_back(next_fib);↵
a = b;↵
b = next_fib;↵
}↵
}↵
↵
// Top-Down DP: count ways to factorize N using fibs[idx..end]↵
long long dp(long long N, int idx) {↵
// Base case: N reduced to 1 = found 1 valid factorization↵
if (N == 1) return 1;↵
↵
// No Fibonacci numbers left but N > 1 = invalid path↵
if (idx >= fibs.size()) return 0;↵
↵
// Check memoization↵
if (memo.count({N, idx})) return memo[{N, idx}];↵
↵
long long ways = 0;↵
↵
// Option 1: Don't use fibs[idx], move to next Fibonacci↵
ways += dp(N, idx + 1);↵
↵
// Option 2: Use fibs[idx] if it divides N↵
// Keep same idx to allow using this Fibonacci multiple times↵
if (N % fibs[idx] == 0) {↵
ways += dp(N / fibs[idx], idx);↵
}↵
↵
return memo[{N, idx}] = ways;↵
}↵
↵
void solve() {↵
long long N;↵
cin >> N;↵
cout << dp(N, 0) << "\n";↵
}↵
↵
int main() {↵
ios_base::sync_with_stdio(false);↵
cin.tie(NULL);↵
↵
precompute();↵
↵
int t;↵
if (cin >> t) {↵
while (t--) solve();↵
}↵
return 0;↵
}↵
```↵
↵
</spoiler>↵
↵
**solution 2: Greedy + Math** author:[user:phsads,2026-05-25]↵
↵
<spoiler summary="Idea">↵
↵
By **Carmichael's theorem**, large Fibonacci numbers (except 2, 3, 8, 144) contain unique prime factors not found in smaller Fibonacci numbers. ↵
↵
This means:↵
- If N is divisible by a large Fibonacci number, we **must** use it in the factorization↵
- We greedily divide N by large Fibonacci numbers first↵
- After this greedy phase, N becomes a product of powers of 2 and 3 only↵
- We count valid combinations of {2, 3, 8, 144}↵
↵
**Algorithm:**↵
1. Generate Fibonacci numbers, excluding 2, 3, 8, 144 (the "exception" numbers)↵
2. Greedily divide N by large Fibonacci numbers (from largest to smallest)↵
3. Factorize remaining N into $2^p \times 3^q$↵
4. Count ways to form $2^p \times 3^q$ using {2, 3, 8, 144}:↵
— Iterate over count of 144 used (each 144 = $2^4 \times 3^2$)↵
— Remaining power of 2 must be formed using {2, 8}↵
— Number of ways to form remaining power P of 2: $\lfloor P/3 \rfloor + 1$↵
↵
**Time Complexity:** Nearly $O(\log N)$ per query ↵
**Space Complexity:** $O(\text{Fib\_count})$↵
↵
</spoiler>↵
↵
<spoiler summary="Code">↵
↵
```cpp↵
#include <iostream>↵
#include <vector>↵
#include <algorithm>↵
↵
using namespace std;↵
↵
vector<long long> greedy_fibs;↵
↵
void precompute() {↵
long long f[90];↵
f[1] = 1; f[2] = 1;↵
↵
// Generate Fibonacci sequence up to 10^18↵
for (int i = 3; i <= 87; i++) {↵
f[i] = f[i-1] + f[i-2];↵
↵
// Exclude exceptions: F_3=2, F_4=3, F_6=8, F_12=144↵
// These don't have unique prime factors↵
if (i != 3 && i != 4 && i != 6 && i != 12) {↵
greedy_fibs.push_back(f[i]);↵
}↵
}↵
↵
// Sort in descending order for greedy approach↵
sort(greedy_fibs.rbegin(), greedy_fibs.rend());↵
}↵
↵
void solve() {↵
long long N;↵
cin >> N;↵
↵
// Step 1: Greedily divide N by large Fibonacci numbers↵
// (they have unique prime factors, so MUST be used if they divide N)↵
for (long long fib : greedy_fibs) {↵
while (N % fib == 0) {↵
N /= fib;↵
}↵
}↵
↵
// Step 2: Factorize remaining N into 2^p * 3^q↵
long long p = 0, q = 0;↵
while (N % 2 == 0) { p++; N /= 2; }↵
while (N % 3 == 0) { q++; N /= 3; }↵
↵
// If remainder is not 1, it has other prime factors → impossible↵
if (N != 1) {↵
cout << 0 << "\n";↵
return;↵
}↵
↵
// Step 3: Count combinations using {2, 3, 8, 144}↵
// where: 144 = 2^4 * 3^2, 8 = 2^3, 3 = 3^1, 2 = 2^1↵
long long ans = 0;↵
↵
// Try all possible counts of 144↵
for (long long c144 = 0; c144 * 4 <= p && c144 * 2 <= q; c144++) {↵
↵
// Remaining power of 2 after using c144 copies of 144↵
long long P_remaining = p - c144 * 4;↵
↵
// Ways to form P_remaining using 8 (2^3) and 2 (2^1)↵
// This is: how many ways to write P as sum of 3's and 1's↵
ans += (P_remaining / 3) + 1;↵
}↵
↵
cout << ans << "\n";↵
}↵
↵
int main() {↵
ios_base::sync_with_stdio(false);↵
cin.tie(NULL);↵
↵
precompute();↵
↵
int t;↵
if (cin >> t) {↵
while (t--) solve();↵
}↵
return 0;↵
}↵
```↵
↵
</spoiler>↵
↵
---↵
↵
## Complexity Comparison↵
↵
| Approach | Time per Query | Space | Notes |↵
|----------|---|---|---|↵
| DP Memoization | $O(\text{Fib}^2 \log N)$ | $O(\text{Fib} \log N)$ | More intuitive, general DP |↵
| Greedy + Math | $O(\log N)$ | $O(\text{Fib})$ | Faster, uses number theory |↵
↵
## Credits↵
↵
Thanks to [user:MEDAAA,2026-05-25], [user:phsads,2026-05-25], [user:Ahmed-Nawaz,2026-05-25], [user:InfiniteLoops1730,2026-05-25], [user:VladiG,2026-05-25], [user:Metall1cA,2026-05-25], [user:helpT27,2026-05-25], and others for all the discussions and ideas in the comments. Really appreciate the different approaches and insights — from DP attempts to number theory explanations.
