/* C05RD_P0W_F C++ Header Example Program.
 *
 * Copyright 2021 Numerical Algorithms Group.
 * Mark 27.2, 2021.
 */

#include <iostream>
#include <math.h>
#include <nag.h>
#include <nagad.h>
#include <nagx02.h>
#include <nagx04.h>
#include <stdio.h>
#include <string>
using namespace std;

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

  cout << "C05RD_P0W_F C++ Header Example Program Results\n\n";

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

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

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

  // Call passive routine
  double diag[n], fjac[n * n], factor, fvec[n], qtf[n], r[n * (n + 1) / 2],
      rwsav[4 * n + 10], xtol;
  Integer irevcm, iwsav[17];

  xtol   = sqrt(X02AJC);
  factor = 100.0;
  for (int i = 0; i < n; ++i)
    {
      diag[i] = 1.0;
    }

  Integer ifail = 0, mode = 2;
  void *  ad_handle = 0;
  irevcm            = 0;
  do
    {
      nag::ad::c05rd(ad_handle, irevcm, n, x, fvec, fjac, xtol, mode, diag,
                     factor, r, qtf, iwsav, rwsav, ifail);

      switch (irevcm)
        {
        case 1:
          // Monitoring exit
          continue;
        case 2:
          for (int i = 0; i < n; ++i)
            {
              fvec[i] = (ruser[1] + ruser[2] * x[i]) * x[i] - ruser[4];
            }
          for (int i = 1; i < n; ++i)
            {
              fvec[i] = fvec[i] + ruser[0] * x[i - 1];
            }
          for (int i = 0; i < n - 1; ++i)
            {
              fvec[i] = fvec[i] + ruser[3] * x[i + 1];
            }
          break;
        case 3:
          for (int i = 0; i < n * n; ++i)
            {
              fjac[i] = 0.0;
            }
          fjac[0] = ruser[1] + 2.0 * ruser[2] * x[0];
          fjac[n] = ruser[3];
          for (int i = 1; i < n - 1; ++i)
            {
              int k       = i * n + i;
              fjac[k - n] = ruser[0];
              fjac[k]     = ruser[1] + 2.0 * ruser[2] * x[i];
              fjac[k + n] = ruser[3];
            }
          fjac[n * n - n - 1] = ruser[0];
          fjac[n * n - 1]     = ruser[1] + 2.0 * ruser[2] * x[n - 1];
          break;
        }
    }
  while (irevcm != 0);

  cout.setf(ios::scientific, ios::floatfield);
  cout.precision(3);
  cout << "           Solution:\n";
  for (int i = 0; i < n; ++i)
    {
      cout.width(8);
      cout << i + 1;
      cout.width(18);
      cout << x[i] << endl;
    }

  return exit_status;
}