Feat: hw2 done

HW2_111550013/HW2_111550013_code/problem3.cpp

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+// 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;
+}