`c- b xx d = b xx d - a`

`c + a = b xx d + b xx d`

`c + a =  2 xx b xx d`

`d = (c + a)/(2 xx b)` (do `b >=1`)

_

#include <bits/stdc++.h>
using namespace std;
long long solve(long long a, long long b, long long c)
{
    return (c+a)/(2*b);
}
int main()
{
    long long a,b,c;
    cin >>a >>b  >> c;
    cout << solve(a,b,c);
    return 0;
}