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: Wrong Answer)

#include <stdio.h>


int main (void){
  
 long long int a=0,b=0,c,max=0,n=0,i,k,flag=0,sum=0,count;
 
 scanf("%lld%lld%lld",&a,&b,&c);
 sum=c;
 count=0;
 for(sum=c;sum<a;count++)
 {
     sum+=b;
     sum+=c;
    
 }
 printf("%lld",(sum>n)?(count-1):count);

return 0;
    
}