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: Accepted)

#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <numeric>
#include <stdio.h>
#include <stack>
#include <queue>
#include <cstdio>
#include <cmath>
#include <iterator>
#include <map>

using namespace std;

#define Int int64_t

void printVec(std::vector<int>& vec) {
	std::cout << "";
	for (auto it = vec.begin(); it != vec.end(); ++it) {
		std::cout << *it << " ";
	}
	std::cout << std::endl;
}
void printVec1(std::vector<string>& vec) {
	std::cout << "";
	for (auto it = vec.begin(); it != vec.end(); ++it) {
		std::cout << *it << " ";
	}
	std::cout << std::endl;
}

template<typename T>
void remove(std::vector<T>& vector, unsigned int index)
{
	vector.erase(vector.begin() + index);
}




int main() {
	int x, y, z; cin >> x>> y>> z;
	int cnt = 0;

	for (int i = 1; y * i + z * (i + 1) <= x; i++) {
		cnt++;
	}

	cout << cnt << endl;

	return 0;
}