/* C05RB_P0W_F C++ Header Example Program.
 *
 * Copyright 2022 Numerical Algorithms Group.
 * Mark 28.3, 2022.
 */

#include <iostream>
#include <math.h>
#include <nag.h>
#include <nagad.h>
#include <nagx02.h>
#include <stdio.h>
#include <string>
using namespace std;

int main()
{
  // Scalars
  int           exit_status = 0;
  const Integer n           = 7;

  cout << "C05RB_P0W_F C++ Header Example Program Results\n\n";

  // problem parameters and starting value
  double r[5], x[7];

  r[0] = -1.0;
  r[1] = 3.0;
  r[2] = -2.0;
  r[3] = -2.0;
  r[4] = -1.0;

  for (int i = 0; i < n; ++i)
  {
    x[i] = -1.0;
  }

  Integer           ifail = 0;
  nag::ad::handle_t ad_handle;
  double            fvec[n], fjac[n * n], xtol;

  xtol = sqrt(X02AJC);
  auto fcn = [&](nag::ad::handle_t &     ad_handle,
                          const Integer &         n,
                          const double *x,
                          double *fvec,
                          double *fjac,
                          Integer &               iflag)
              {
                if (iflag != 2)
                {
                  for (int i = 0; i < n; ++i)
                  {
                    fvec[i] = (r[1] + r[2] * x[i]) * x[i] - r[4];
                  }
                  for (int i = 1; i < n; ++i)
                  {
                    fvec[i] = fvec[i] + r[0] * x[i - 1];
                  }
                  for (int i = 0; i < n - 1; ++i)
                  {
                    fvec[i] = fvec[i] + r[3] * x[i + 1];
                  }
                }
                else
                {
                  for (int i = 0; i < n * n; ++i)
                  {
                    fjac[i] = 0.0;
                  }
                  fjac[0] = r[1] + 2.0 * r[2] * x[0];
                  fjac[n] = r[3];
                  for (int i = 1; i < n - 1; ++i)
                  {
                    int k       = i * n + i;
                    fjac[k - n] = r[0];
                    fjac[k]     = r[1] + 2.0 * r[2] * x[i];
                    fjac[k + n] = r[3];
                  }
                  fjac[n * n - n - 1] = r[0];
                  fjac[n * n - 1]     = r[1] + 2.0 * r[2] * x[n - 1];
                }
                iflag = 0;
              };

  nag::ad::c05rb(ad_handle, fcn, n, x, fvec, fjac, xtol, ifail);

  cout.setf(ios::scientific, ios::floatfield);
  cout.precision(4);
  cout << "           Solution:\n";
  for (int i = 0; i < n; ++i)
  {
    cout.width(10);
    cout << i + 1;
    cout.width(20);
    cout << x[i] << endl;
  }

  return exit_status;
}