Последовательность Фибоначи

#pragma once
#define _CRT_SECURE_NO_WARNINGS
#define _USE_MATH_DEFINES
#include <iostream>
#include <string>
#include <map>
#include <set>
#include <algorithm>
#include <vector>
#include <stdio.h>
#include <cmath>
#include <math.h>
#include <queue>
#include <stack>
#include <climits>
#include <deque>
#include <ctime>
#include <iomanip>
#include <bitset>
#include <unordered_map>
#include <unordered_set>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int ui;

#define mh() make_heap()
#define poph() pop_heap()
#define pushh() push_heap()
#define sor(n) n.begin(), n.end()
#define rsor(n) n.rbegin(), n.rend()
#define mp make_pair
#define files freopen("input.txt", "rt", stdin); freopen("output.txt", "wt", stdout)
#define p(T) pair<T,T>
#define toch(x) cout.precision(x), cout.setf(ios::fixed)
#define znac(l) abs(l)/l
#define IOS ios::sync_with_stdio(false)
#define IOSB cin.tie(0), cout.tie(0);
const ll ok = ll(1e9 + 7);

int main()
{
    IOSB;
    IOS;
#ifdef TheCompiler
    files;
#endif
    toch(0);
    ll i, j, n;
    vector<long double>f(2010);
    cin >> i ;
    i += 1000;
    cin >> f[i];
    cin >> j;
    j += 1000;
    cin >> f[j] >> n;
    n += 1000;
    if (i > j)
    {
        swap(i, j);
    }
    if (i == n)
    {
        cout << f[i];
        return 0;
    }
    if (j == n)
    {
        cout << f[j];
        return 0;
    }
    ll sc = 0;
    if ((j - i - 1) % 2 == 0)
    {
        sc = i - 1;
    }
    else
    {
        sc = i+1;
    }
    vector<p(ll)>ck(2010);
    ck[sc] = { 0,1 };
    ck[i] = { 1,0 };
    for (int k= max(sc, i)+1; k <= j; k++)
    {
        ck[k].first = ck[k - 1].first + ck[k - 2].first;
        ck[k].second = ck[k - 1].second + ck[k - 2].second;
    }
    f[sc] = (f[j] - ck[j].first*f[i]) / ck[j].second;
    if (n >= min(sc, i))
    {
        for (int k = max(sc, i) + 1; k <= n; k++)
        {
            f[k] = f[k - 1] + f[k - 2];
        }
        cout << f[n];
    }
    else
    {
        for (int k = min(sc, i) - 1; k >= n; k--)
        {
            f[k] = f[k - 2] - f[k - 1];
        }
        cout << f[n];
    }
    return 0;
}