33 lines
893 B
C++
33 lines
893 B
C++
// Problem 3: Symmetric tridiagonal system
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#include <iostream>
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#include <iomanip>
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#include <vector>
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using namespace std;
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int main() {
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int n = 6;
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vector<double> d = {4, 4, 4, 4, 4, 4}; // diagonal
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vector<double> a = {-1, -1, -1, -1, -1}; // off-diagonal
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vector<double> b = {100, 200, 200, 200, 200, 100};
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// Forward elimination
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for (int i = 1; i < n; i++) {
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double m = a[i-1] / d[i-1];
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d[i] -= m * a[i-1];
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b[i] -= m * b[i-1];
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}
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// Back substitution
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vector<double> x(n);
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x[n-1] = b[n-1] / d[n-1];
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for (int i = n - 2; i >= 0; i--)
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x[i] = (b[i] - a[i] * x[i+1]) / d[i];
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cout << "Solution:\n";
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for (int i = 0; i < n; i++)
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cout << "x" << (i+1) << " = " << fixed << setprecision(6) << x[i] << "\n";
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cout << "\nOperations: 7N - 6 = " << (7*n - 6) << " for N = " << n << "\n";
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return 0;
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}
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