```/* C05RD_P0W_F C++ Header Example Program.
*
* Copyright 2019 Numerical Algorithms Group.
* Mark 27, 2019.
*/

#include <nag.h>
#include <stdio.h>
#include <math.h>
#include <nagx02.h>
#include <nagx04.h>
#include <iostream>
#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;
irevcm = 0;
do {
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;
}
```