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"bits/stdc++.h"
using namespace std;

#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define rep3(i,m,n) for(int (i)=m;(i)<=(n);(i)++)
#define rep3rev(i,m,n) for(int (i)=m;(i)>=(n);(i)--)
#define all(a) (a.begin()),(a.end())
#define rall(a) (a.rbegin()),(a.rend())
#define fi first
#define se second

typedef long long ll;
typedef vector<ll> vll;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef pair<int, int> pii;

void Main()
{
  	int x, y, z;
  	cin >> x >> y >> z;
  	cout << (x-1)/(y+z) << endl;
	return;
}

int main()
{
	cin.tie(nullptr);
	ios_base::sync_with_stdio(false);
	cout << fixed << setprecision(15);
	Main();
	return 0;
}