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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef vector<vector<int>> Graph;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
#define pi 3.14159265359
#define inf 2147483647
#define INF 9223372036854775807
#define mod 1000000007
#define mod2 998244353

int main() {
    int X, Y, Z; cin >> X >> Y >> Z;
    ll sum = 0, cnt = 0;
    if(2 * Z + Y < X) {
        cnt++;
        sum += 2 * Z + Y;
        while(sum + Y + Z <= X) {
            sum += Z + Y;
            cnt++;
        }
    }
    cout << cnt << endl;
    return 0;
}