// Problem 2: Scaled partial pivoting
#include <iostream>
#include <iomanip>
#include <cmath>
#include <vector>
using namespace std;

double roundSig(double val, int sig) {
    if (val == 0) return 0;
    double d = ceil(log10(fabs(val)));
    double p = pow(10, sig - d);
    return round(val * p) / p;
}

void solve(vector<vector<double>> A, int n, int sig) {
    cout << "\n" << sig << " significant digits:\n";

    vector<double> s(n);
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++) s[i] = max(s[i], fabs(A[i][j]));

    vector<int> idx(n);
    for (int i = 0; i < n; i++) idx[i] = i;

    for (int k = 0; k < n - 1; k++) {
        int maxRow = k;
        double maxRatio = fabs(A[idx[k]][k]) / s[idx[k]];
        for (int i = k + 1; i < n; i++) {
            double ratio = fabs(A[idx[i]][k]) / s[idx[i]];
            if (ratio > maxRatio) { maxRatio = ratio; maxRow = i; }
        }
        swap(idx[k], idx[maxRow]);

        for (int i = k + 1; i < n; i++) {
            double m = roundSig(A[idx[i]][k] / A[idx[k]][k], sig);
            for (int j = k; j <= n; j++)
                A[idx[i]][j] = roundSig(A[idx[i]][j] - m * A[idx[k]][j], sig);
        }
    }

    vector<double> x(n);
    for (int i = n - 1; i >= 0; i--) {
        x[i] = A[idx[i]][n];
        for (int j = i + 1; j < n; j++)
            x[i] = roundSig(x[i] - A[idx[i]][j] * x[j], sig);
        x[i] = roundSig(x[i] / A[idx[i]][i], sig);
    }

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

int main() {
    vector<vector<double>> A = {
        {4.13, -2.20, 0.95, 3.02},
        {6.14, 4.45, -1.45, -4.02},
        {1.03, 1.86, 0.44, 5.22}
    };
    solve(A, 3, 6);
    solve(A, 3, 3);
    return 0;
}