For this Question: Ready for a super simple challenge!
You are given two integers A and B. You will create a line of boys and girls in the following manner:
First A girls, then B boys, then A girls, then B boys, and so on.
You will also be given an integer N. Find the number of girls in the first N people of the row.The first and the only line of input contains 3 integers N, A, B.

Constraints
1 < N, A, B <= 10^18 (1000000000000000000)
A+B <= 10^18Output a singer integer, the number of girls in the first N people in the queue.Sample Input
8 3 4

Sample Output
4

Explanation
The arrangement will look like GGGBBBBG, where G represents girl and B represents boy., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define int long long
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define MOD 1000000007
#define INF 1000000000000000007LL

const int N = 100005;
 
// it's swapnil07 ;)
 
#ifdef SWAPNIL07
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define end_routine()
#endif


signed main()
{
    fast
  
    int n, a, b; cin>>n>>a>>b;
    int x = a+b;
    int y = (n/x);
    n -= (y*x);
    int ans = y*a;
    if(n > a)
        ans += a;
    else
        ans += n;
    cout<<ans;
    return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not?