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<cstdio>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<string>
#include<cstring>
using namespace std;
int main(){
	int x,y,z,sum,flag;
	while(~scanf("%d%d%d",&x,&y,&z)){
		sum=0;
		flag=0;
		for(int i=1; ;i++){
			sum = sum+z;
			if(sum>=x) break;
			sum = sum+y;
			if(sum>x-z) break;
			flag++;
		}
		printf("%d\n",flag);
	}
}