Problem:
Score : 200 points

Problem Statement

We have a long seat of width X centimeters.
There are many people who wants to sit here. A person sitting on the seat will always occupy an interval of length Y centimeters.

We would like to seat as many people as possible, but they are all very shy, and there must be a gap of length at least Z centimeters between two people, and between the end of the seat and a person.

At most how many people can sit on the seat?

Constraints

All input values are integers.

1 \leq X, Y, Z \leq 10^5

Y+2Z \leq X

Input

Input is given from Standard Input in the following format:

X Y Z

Output

Print the answer.

Sample Input 1

13 3 1

Sample Output 1

3

There is just enough room for three, as shown below:

Figure

Sample Input 2

12 3 1

Sample Output 2

2

Sample Input 3

100000 1 1

Sample Output 3

49999

Sample Input 4

64146 123 456

Sample Output 4

110

Sample Input 5

64145 123 456

Sample Output 5

109

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

#include <iostream>
#include <fstream>
#include <sstream>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <unordered_set>
#include <unordered_map>
#include <bitset>
#include <limits>
#include <random>
#include <complex>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <string.h>
#include <iomanip>
using namespace std;
  
#define REP(i,n) for(int (i)=0;(i)<(int)(n);(i)++)
#define RREP(i,n) for(int (i)=(int)(n)-1;i>=0;i--)
#define REMOVE(Itr,n) (Itr).erase(remove((Itr).begin(),(Itr).end(),n),(Itr).end())
typedef long long ll;

int main() {

    ll X,Y,Z; cin >> X >> Y >> Z;
    X -= Z * 2;
    cout << (X / (Z + Y)) << endl;

    return 0;
}