高斯消元

P3389 【模板】高斯消元法 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

/*
 * @Author: zhl
 * @Date: 2020-11-14 09:22:58
 */
#include<bits/stdc++.h>
using namespace std;

double A[110][110];
int n;
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n + 1; j++) {
			scanf("%lf", &A[i][j]);
		}
	}
	for (int i = 1; i <= n; i++) {
		int mx = i;
		for (int j = i + 1; j <= n; j++) {
			if (fabs(A[j][i]) > fabs(A[mx][i])) mx = j;
		}

		for (int j = 1; j <= n + 1; j++)swap(A[i][j], A[mx][j]);
		if (fabs(A[i][i]) < 1e-8) {
			puts("No Solution");
			return 0;
		}
		for (int j = 1; j <= n; j++) {
			if (i == j)continue;
			double t = A[j][i] / A[i][i];
			for (int k = i + 1; k <= n + 1; k++) {
				A[j][k] -= A[i][k] * t;
			}
		}
	}
	for (int i = 1; i <= n; i++) {
		printf("%.2f
", A[i][n + 1] / A[i][i]);
	}
}