// Problem 3: Symmetric tridiagonal system
#include <iostream>
#include <iomanip>
#include <vector>
using namespace std;

int main() {
    int n = 6;
    vector<double> d = {4, 4, 4, 4, 4, 4};      // diagonal
    vector<double> a = {-1, -1, -1, -1, -1};    // off-diagonal
    vector<double> b = {100, 200, 200, 200, 200, 100};

    // Forward elimination
    for (int i = 1; i < n; i++) {
        double m = a[i-1] / d[i-1];
        d[i] -= m * a[i-1];
        b[i] -= m * b[i-1];
    }

    // Back substitution
    vector<double> x(n);
    x[n-1] = b[n-1] / d[n-1];
    for (int i = n - 2; i >= 0; i--)
        x[i] = (b[i] - a[i] * x[i+1]) / d[i];

    cout << "Solution:\n";
    for (int i = 0; i < n; i++)
        cout << "x" << (i+1) << " = " << fixed << setprecision(6) << x[i] << "\n";

    cout << "\nOperations: 7N - 6 = " << (7*n - 6) << " for N = " << n << "\n";
    return 0;
}