Доп. из 10.2. Простые вычисления

#include<iostream>
#include<cmath>
using namespace std;
int N;
const double eps = 1e-9;
double a[3010], c[3010], x;
void solve()
{
    for (int i = 2; i <= N; i++)a[i] = 2 * (a[i - 1] + c[i - 1]) - a[i - 2];
    x = 2 * (a[N] + c[N]) - a[N - 1];
}
int main()
{
    scanf("%d", &N);
    scanf("%lf%lf", &a[0], &a[N + 1]);
    for (int i = 1; i <= N; i++)cin >> c[i];
    double l = -2000.000, r = 2000.00;
    while (l - r<eps)
    {
        double mid = (l + r) / 2.0;
        a[1] = mid;
        solve();
        if (fabs(a[N + 1] - x)<eps){ printf("%.2lf\n", mid); break; }
        else if (a[N + 1] - x<eps)r = mid; else l = mid;
    }
    return 0;
}