模板如下,复杂度O(n^3)


/* 高斯消元低精度版 @挠头小熊熊
 *
 *	使用说明:
 *		A为大小为n的增广矩阵,A[i][n]是第i个方程右边的常数bi
 *		运行后A[i][n]是第i个未知数的值
 *	注意:
 *		函数第一个形参的第二维大小记得更改
 *		本版精度不高
 */
#include <cstdio>
#include <algorithm>
using namespace std;

void gs(double A[][5], int n) {
	int i, j, k, r;
	for (i = 0; i < n; i++) {
		r = i;
		for (j = i + 1; j < n; j++)
			if (fabs(A[j][i]) > fabs(A[r][i])) r = j;
		if (r != i) for (j = 0; j <= n; j++) swap(A[r][j], A[i][j]);

		for (j = n; j >= i; j--)
			for (k = i + 1; k < n; k++)
				A[k][j] -= A[k][i] / A[i][i] * A[i][j];
	}
	for (i = n - 1; i >= 0; i--) {
		for (j = i + 1; j < n; j++)
			A[i][n] -= A[j][n] * A[i][j];
		A[i][n] /= A[i][i];
	}
}

double A[][5] = {{2, 1, -1, 8}, { -3, -1, 2, -11}, { -2, 1, 2, -3}};

int main() {
	//freopen("in.in", "r", stdin);
	//freopen("out.out", "w", stdout);
	gs(A, 3);
	for (int i = 0; i < 3; i++)
		printf("%.3f\n", A[i][3]);
	return 0;
}