备战NOIP——模板复习3

这里只有模板，并不作讲解，仅为路过的各位做一个参考以及用做自己复习的资料，转载注明出处。

高斯消元

/*Copyright: Copyright (c) 2018
*Created on 2018-10-28  
*Author: 十甫
*Version 1.0 
*Title: 高斯消元
*Time: inf mins
*/ 
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
const int size = 105;
const double eps = 1e-7;

int n;
double matrix[size][size], ans[size];
inline bool Gauss() {
	for(int i = 1;i <= n;i++) {
		int maxNumLine = i;
		for(int j = i + 1;j <= n;j++)
			if(fabs(matrix[maxNumLine][i]) < fabs(matrix[j][i]))
				maxNumLine = j;
		if(fabs(matrix[maxNumLine][i]) < eps)
			return false;
		if(i != maxNumLine) 
			swap(matrix[i], matrix[maxNumLine]);
		double div = matrix[i][i];
		for(int j = i;j <= n + 1;j++)
			matrix[i][j] /= div;
		for(int j = i + 1;j <= n;j++) {
			div = matrix[j][i];
			for(int k = i;k <= n + 1;k++) {
				matrix[j][k] -= matrix[i][k] * div;
			}
		}
	}
	ans[n] = matrix[n][n + 1];
	for(int i = n - 1;i >= 1;i--) {
		ans[i] = matrix[i][n + 1];
		for(int j = i + 1;j <= n;j++) {
			ans[i] -= (matrix[i][j] * ans[j]);
		}
	}
	return true;
}

int main() {
	scanf("%d", &n);
	for(int i = 1;i <= n;i++) {
		for(int j = 1;j <= n + 1;j++) {
			scanf("%lf", &matrix[i][j]);
		}
	}
	if(!Gauss()) printf("No Solution
");
	else for(int i = 1;i <= n;i++) {
		printf("%.2lf
", ans[i]);
	}
	return 0;
}