Problem:
Score : 200 points

Problem Statement

There are N apple trees in a row. People say that one of them will bear golden apples.

We want to deploy some number of inspectors so that each of these trees will be inspected.

Each inspector will be deployed under one of the trees. For convenience, we will assign numbers from 1 through N to the trees. An inspector deployed under the i-th tree (1 \leq i \leq N) will inspect the trees with numbers between i-D and i+D (inclusive).

Find the minimum number of inspectors that we need to deploy to achieve the objective.

Constraints

All values in input are integers.

1 \leq N \leq 20

1 \leq D \leq 20

Input

Input is given from Standard Input in the following format:

N D

Output

Print the minimum number of inspectors that we need to deploy to achieve the objective.

Sample Input 1

6 2

Sample Output 1

2

We can achieve the objective by, for example, placing an inspector under Tree 3 and Tree 4.

Sample Input 2

14 3

Sample Output 2

2

Sample Input 3

20 4

Sample Output 3

3

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

#ifndef bhartiya
    #pragma GCC optimize("Ofast")
    #pragma GCC optimize("unroll-loops")
    // #pragma GCC target ("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#endif

#define int long long
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef gp_hash_table<int, int> umap;
typedef tree<int, null_type, less<int>, rb_tree_tag,
		tree_order_statistics_node_update> oset;
// not that imp
typedef pair<pii, int> piii;
typedef vector<int> vi;
typedef vector<pii> vii;

#define inf 200000000000000
#define mod 1000000007
#define eps 1e-7
#define PI 3.1415926535897932385
// #define PI acos(-1)

#define pb push_back
#define bitc  __builtin_popcountll
#define mp make_pair
#define ff first
#define ss second
#define all(ar) ar.begin(), ar.end()
#define sz(x) (int)(x).size()

#define fr(i,a,b) for (int i = (a), _b = (b); i <= _b; i++)
#define rep(i,n) for (int i = 0, _n = (n); i < _n; i++)
#define repr(i,n) for (int i = n - 1; i >= 0; i--)
#define frr(i,a,b) for (int i = (a), _b = (b); i >= _b; i--)
#define foreach(it,ar) for (auto it = ar.begin(); it != ar.end(); it++)
#define fil(ar,val) memset(ar, val, sizeof(ar))

#ifdef bhartiya
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
	cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
	const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif

mt19937 rng32(chrono::steady_clock::now().time_since_epoch().count());
inline bool equals(double a, double b) {return fabs(a - b) < 1e-9;}
int gcd(int a, int b){ return b==0 ? a : gcd(b, a%b); }
ll modpow(ll b, ll e) {
	ll ans=1; for(;e;b=b*b%mod,e/=2) if(e&1) ans=ans*b%mod; return ans;}

const int maxn = 1000005;

void solve(){
	ld n,d; cin>>n>>d;
	int ans = ceil(n/(2*d + 1));
	cout<<ans;
}

signed main() 
{ 
	ios_base::sync_with_stdio(false);
    cin.tie(0), cout.tie(0), cout.precision(10); //cout<<fixed;
    cin.exceptions(cin.failbit);
    #ifdef bhartiya
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
    #endif
    solve();
    end_routine();
}