线性方程组(System of Linear Equations) 题解

思路分析

按题目要求使用对应数据结构与算法模拟/递推求解，注意数据范围边界与时间复杂度。

参考代码

#include <cmath>
#include <iomanip>
#include <iostream>
#include <vector>
using namespace std;

const double EPS = 1e-8;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    if (!(cin >> n)) return 0;
    vector<vector<double>> a(n, vector<double>(n + 1));
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j <= n; ++j) cin >> a[i][j];
    }

    int r = 0;
    for (int col = 0; col < n && r < n; ++col) {
        int piv = r;
        for (int i = r + 1; i < n; ++i) {
            if (fabs(a[i][col]) > fabs(a[piv][col])) piv = i;
        }
        if (fabs(a[piv][col]) < EPS) continue;
        swap(a[r], a[piv]);
        for (int i = 0; i < n; ++i) {
            if (i != r && fabs(a[i][col]) > EPS) {
                double factor = a[i][col] / a[r][col];
                for (int j = col; j <= n; ++j) a[i][j] -= factor * a[r][j];
            }
        }
        r++;
    }

    for (int i = r; i < n; ++i) {
        if (fabs(a[i][n]) > EPS) {
            cout << -1 << "\n";
            return 0;
        }
    }
    if (r < n) {
        cout << 0 << "\n";
        return 0;
    }

    cout << fixed << setprecision(2);
    for (int i = 0; i < n; ++i) {
        double val = a[i][n] / a[i][i];
        if (fabs(val) < 0.005) val = 0.0;
        cout << "x" << (i + 1) << "=" << val << "\n";
    }
    return 0;
}