每次寻找同一列最大系数所在的行数并交换到与列数相等的行数。后用正常解方程思想消掉这一列的系数只剩下最大系数的主元

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>

using namespace std;

double a[11451][11451]; 
int n;

inline double find_pivot( int column /*列数*/) {
	//一列中最大的系数为主元 
	//并寻找这一列主元所在的行数 
	int maxn = column;
	for (int i = column + 1; i <= n; ++i) {
		if ( fabs( a[i][column] ) > fabs( a[maxn][column] ) ) {
			maxn = i;
		}
	} 
	return maxn;
}

inline void swap_line( int beg , int end ) {
	//让beg行交换主元最大的end行 
	for (int i = 1; i <= n + 1; ++i) {
		swap( a[beg][i] , a[end][i] );
	}
	return;
}

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) {
		double pivot = find_pivot ( i );//主元所在的行 
		swap_line ( i , pivot ); 
		if ( a[i][i] == 0 ) {
			//存在一列的系数均为0
			//则这一列的未知数一定无解 
			printf("No Solution\n");
			return 0; 
		} 
		for (int j = 1; j <= n; ++j) {
			if ( j != i ) {
				double temp = a[j][i] / a[i][i];
				//算出第j行的第i列与主元的倍数关系
				for (int k = i; k <= n + 1; ++k) {
					a[j][k] -= ( a[i][k] * temp );
					//消去与主元同一列的系数 
				} 
			}
		}
	} 
	for (int i = 1; i <= n; ++i) {
		a[i][n + 1] /= a[i][i];
		//将最后的结果除以第i项的系数 
	}
	for (int i = 1; i <= n; ++i) {
		printf("%.2lf\n",a[i][n + 1]);
	}
	return 0;
}