【模板】高斯消元法

题目描述:https://www.luogu.org/problem/P3389

#include<bits/stdc++.h>
using namespace std;
int n;
double a[105][105];
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 pi=i;
		while(!a[pi][i]&&pi<=n) pi++;
		if(pi>n){
			printf("No Solution
");
			return 0;
		}
		for(int j=1;j<=n+1;j++) 
			swap(a[i][j],a[pi][j]);
		double t=a[i][i];
		for(int j=1;j<=n+1;j++) 
			a[i][j]/=t;
		for(int j=1;j<=n;j++){
			if(j!=i){
				double t=a[j][i];
				for(int m=1;m<=n+1;m++)
					a[j][m]-=t*a[i][m];
			}
		}
	}
	for(int i=1;i<=n;i++) 
		printf("%.2lf
",a[i][n+1]);
	return 0;
} 
原文地址:https://www.cnblogs.com/zhangdading/p/11484241.html