Problem:
Score : 800 points

Problem Statement

Snuke has a string x of length N.
Initially, every character in x is 0.

Snuke can do the following two operations any number of times in any order:

Choose A consecutive characters in x and replace each of them with 0.

Choose B consecutive characters in x and replace each of them with 1.

Find the number of different strings that x can be after Snuke finishes doing operations.
This count can be enormous, so compute it modulo (10^9+7).

Constraints

1 \leq N \leq 5000

1 \leq A,B \leq N

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N A B

Output

Print the number of different strings that x can be after Snuke finishes doing operations, modulo (10^9+7).

Sample Input 1

4 2 3

Sample Output 1

11

For example, x can be 0011 or 1111 in the end, but cannot be 0110.

Sample Input 2

10 7 2

Sample Output 2

533

Sample Input 3

1000 100 10

Sample Output 3

828178524

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

template <int MOD_> struct modnum {
	static constexpr int MOD = MOD_;
	static_assert(MOD_ > 0, "MOD must be positive");

private:
	using ll = long long;

	int v;

	static int minv(int a, int m) {
		a %= m;
		assert(a);
		return a == 1 ? 1 : int(m - ll(minv(m, a)) * ll(m) / a);
	}

public:

	modnum() : v(0) {}
	modnum(ll v_) : v(int(v_ % MOD)) { if (v < 0) v += MOD; }
	explicit operator int() const { return v; }
	friend std::ostream& operator << (std::ostream& out, const modnum& n) { return out << int(n); }
	friend std::istream& operator >> (std::istream& in, modnum& n) { ll v_; in >> v_; n = modnum(v_); return in; }

	friend bool operator == (const modnum& a, const modnum& b) { return a.v == b.v; }
	friend bool operator != (const modnum& a, const modnum& b) { return a.v != b.v; }

	modnum inv() const {
		modnum res;
		res.v = minv(v, MOD);
		return res;
	}
	friend modnum inv(const modnum& m) { return m.inv(); }
	modnum neg() const {
		modnum res;
		res.v = v ? MOD-v : 0;
		return res;
	}
	friend modnum neg(const modnum& m) { return m.neg(); }

	modnum operator- () const {
		return neg();
	}
	modnum operator+ () const {
		return modnum(*this);
	}

	modnum& operator ++ () {
		v ++;
		if (v == MOD) v = 0;
		return *this;
	}
	modnum& operator -- () {
		if (v == 0) v = MOD;
		v --;
		return *this;
	}
	modnum& operator += (const modnum& o) {
		v += o.v;
		if (v >= MOD) v -= MOD;
		return *this;
	}
	modnum& operator -= (const modnum& o) {
		v -= o.v;
		if (v < 0) v += MOD;
		return *this;
	}
	modnum& operator *= (const modnum& o) {
		v = int(ll(v) * ll(o.v) % MOD);
		return *this;
	}
	modnum& operator /= (const modnum& o) {
		return *this *= o.inv();
	}

	friend modnum operator ++ (modnum& a, int) { modnum r = a; ++a; return r; }
	friend modnum operator -- (modnum& a, int) { modnum r = a; --a; return r; }
	friend modnum operator + (const modnum& a, const modnum& b) { return modnum(a) += b; }
	friend modnum operator - (const modnum& a, const modnum& b) { return modnum(a) -= b; }
	friend modnum operator * (const modnum& a, const modnum& b) { return modnum(a) *= b; }
	friend modnum operator / (const modnum& a, const modnum& b) { return modnum(a) /= b; }
};

template <typename T> T pow(T a, long long b) {
	assert(b >= 0);
	T r = 1; while (b) { if (b & 1) r *= a; b >>= 1; a *= a; } return r;
}

using num = modnum<int(1e9) + 7>;

vector<num> fact, ifact;

void init(){
	int N = 1100000;
	fact = {1};
	for(int i = 1; i < N; i++) fact.push_back(i * fact[i-1]);
	ifact.resize(N);
	ifact.back() = 1 / fact.back();
	for(int i = N - 1; i > 0; i--) ifact[i-1] = i * ifact[i];
}

num ncr(int n, int k){
	if(k < 0 || k > n) return 0;
	return fact[n] * ifact[k] * ifact[n-k];
}

using ll = long long;

int main(){
	ios_base::sync_with_stdio(false), cin.tie(nullptr);
	init();
	int n, a, b;
	cin >> n >> a >> b;
	if(a > b) swap(a, b);
	vector<vector<num> > waysb(n+1, vector<num>(n+1, 0));
	vector<vector<num> > waysa(n+1, vector<num>(n+1, 0));
	waysb[0][0] = 1;
	vector<num> bada(n+1, 0);
	for(int i = 0; i < n; i++){
		num totalb = 0;
		for(int j = 0; j < b; j++) totalb += waysb[i][j];
		for(int j = b; j <= n; j++){
			waysa[i+1][j] += waysa[i][j] * 2;
			waysb[i+1][j] += waysb[i][j] * 2;
		}
		for(int j = 0; j <= i && j < b; j++){
			waysa[i+1][j+1] += waysa[i][j];
			waysb[i+1][j+1] += waysa[i][j];
			waysb[i+1][j+1] += waysb[i][j];
		}
		waysb[i+1][1] += bada[i];
		if(i + a <= n){
			for(int j = 0; j <= i && j < b; j++){
				waysa[i+a][j+a] += waysb[i][j];
			}			
		}
		for(int ri = i+1; ri < i+a && ri <= n; ri++){
			bada[ri] += totalb;
		}
	}
	num ans = 0;
	for(int j = b; j <= n; j++){
		ans += waysa[n][j];
		ans += waysb[n][j];
	}
	cout << ans << '\n';
}