注意事项：

  1. 高斯消元代码的 14 行和 17~20 行也可向矩阵行列式那样包装成一个函数并使用 bottomj 优化常数。但因为在 Luogu P2455 [SDOI2006]线性方程组 的测试中对常数的优化不明显，又因为 swap(a[row], a[maxer]) 在高消中只出现一次，这样做也不会是代码结构更清晰，所以不推荐这种写法。

  2. 该高斯消元代码只适用于矩阵行列长度相等，适用于行列长度不相等的版本请见异或高斯消元。

代码：

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int n;
double a[MAXn + 10][MAXn + 10];
int guass() {
    int row = 1;
    for (int col = 1; col <= n; ++col) {
        double mx = 0; int maxer = -1;
        for (int i = row; i <= n; ++i) {
            if (mx < abs(a[i][col])) {
                mx = abs(a[i][col]);
                maxer = i;
            }
        }
        if (sig(mx) == 0) continue;
        if (maxer != row) swap(a[row], a[maxer]);
        for (int i = 1; i <= n; ++i) {
            if (i == row) continue;
            double k = a[i][col] / a[row][col];
            for (int j = col; j <= n + 1; ++j) {
                a[i][j] -= a[row][j] * k;
            }
        }
        ++row;
    }
    bool noans = 0, infans = 0;
    if (row <= n) {
        for (int i = row; i <= n; ++i) {
            if (sig(a[i][n + 1])) {
                noans = 1;
                break;
            }
        }
        if (noans == 0) infans = 1;
    }
    if (noans) return 0;
    else if (infans) return -1;
    else {
        for (int i = 1; i <= n; ++i) {
            a[i][n + 1] /= a[i][i];
            a[i][i] = 1;
        }
        return 1;
    }
}

Luogu P2455 [SDOI2006]线性方程组

代码：

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#include<bits/stdc++.h>
using namespace std;
const int MAXn = 50;
const double EPS = 1e-8;

template <typename T>
inline void read(T &a) {
    char c;for (c = getchar(); (c < '0' || c > '9') && c != '-'; c = getchar());bool f = c == '-';T x = f ? 0 : (c ^ '0');for (c = getchar(); c >= '0' && c <= '9'; c = getchar()) {x = x * 10 + (c ^ '0');}a = f ? -x : x;
}
template <typename T, typename ...Argv>
inline void read(T &a, Argv &...argv) {
    read(a), read(argv...);
}

inline int sig(double x) {
    if (x < -EPS) return -1;
    else if (x > EPS) return 1;
    else return 0;
}

int n;
double a[MAXn + 10][MAXn + 10];
int guass() {
    int row = 1;
    for (int col = 1; col <= n; ++col) {
        double mx = 0; int maxer = -1;
        for (int i = row; i <= n; ++i) {
            if (mx < abs(a[i][col])) {
                mx = abs(a[i][col]);
                maxer = i;
            }
        }
        if (sig(mx) == 0) continue;
        if (maxer != row) swap(a[row], a[maxer]);
        for (int i = 1; i <= n; ++i) {
            if (i == row) continue;
            double k = a[i][col] / a[row][col];
            for (int j = col; j <= n + 1; ++j) {
                a[i][j] -= a[row][j] * k;
            }
        }
        ++row;
    }
    bool noans = 0, infans = 0;
    if (row <= n) {
        for (int i = row; i <= n; ++i) {
            if (sig(a[i][n + 1])) {
                noans = 1;
                break;
            }
        }
        if (noans == 0) infans = 1;
    }
    if (noans) return 0;
    else if (infans) return -1;
    else {
        for (int i = 1; i <= n; ++i) {
            a[i][n + 1] /= a[i][i];
            a[i][i] = 1;
        }
        return 1;
    }
}

signed main() {
    read(n);
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n + 1; ++j) {
            scanf("%lf", &a[i][j]);
        }
    }
    int res = guass();
    if (res == 0) puts("-1");
    else if (res == -1) puts("0");
    else {
        for (int i = 1; i <= n; ++i) {
            printf("x%d=%.2lf\n", i, a[i][n + 1]);
        }
    }
    return 0;
}