#include <iostream>
#include <vector>
#include <string>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <algorithm>	// require sort next_permutation count __gcd reverse etc.
#include <cstdlib>	// require abs exit atof atoi 
#include <cstdio>		// require scanf printf
#include <functional>
#include <numeric>	// require accumulate
#include <cmath>		// require fabs
#include <climits>
#include <limits>
#include <cfloat>
#include <iomanip>	// require setw
#include <sstream>	// require stringstream 
#include <cstring>	// require memset
#include <cctype>		// require tolower, toupper
#include <fstream>	// require freopen
#include <ctime>		// require srand
#define rep(i,n) for(int i=0;i<(n);i++)
#define ALL(A) A.begin(), A.end()
#define INF 1e100

/*
	No.23 技の選択	

	http://yukicoder.me/problems/33

	動的計画法

	dp[i]:= 現在の攻撃力が i のとき 通常攻撃または必殺技で攻撃した場合 最小の期待値
	dp[i] = min (dp[i-A] + 1, dp[i-D] + 3/2 );

	H ≦ i ≦ INF の範囲で res = min (res, dp[i] ) 
	
*/

using namespace std;

typedef long long ll;
typedef pair<int, int> P;

const int MAX_H = 2*10005;
double dp[MAX_H];

int main()
{
	rep (i, MAX_H ) dp[i] = INF;
	ios_base::sync_with_stdio(0);
	int H, A, D; cin >> H >> A >> D;

	dp[0] = 0.;
	for (int i = 0; i < MAX_H; i++ ){
		if (i - A >= 0 )	
			dp[i] = min (dp[i], dp[i-A] + 1. );
		if (i - D >= 0 )
			dp[i] = min (dp[i], dp[i-D] + 3./2. );
	} // end if
	double res = INF;
	for (int i = MAX_H-1; i >= H; i-- )
		res = min (res, dp[i] );
	
	printf ("%.5lf\n", res );

	return 0;
}