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 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <limits.h>
#include <stdbool.h>

typedef double db;
typedef int in;
typedef long int li;
typedef long long int lli;
typedef long ld;
typedef long lld;

#define pi M_PI
#define in scanf
#define out printf
#define inf INT_MAX
#define rep(i,n) for(int i=0;i<n;++i)
#define debugd(d) printf("%d\n",d)
#define debugs(s) printf("%s\n",s)
#define debugc(c) printf("%c\n",c)
#define debugf(f) printf("%f\n",f)

int compare_int(const void *a, const void *b)
{
    return *(int*)a - *(int*)b;
}

int main (void)
{
  int x,y,z;
  in("%d %d %d\n",&x,&y,&z);
  rep(i,100010)
  {
    if(z+i*(y+z)>x)
    {
      printf("%d \n",i-1);
      return 0;
    }
  }
  return 0;
}