理论

高斯消元法，是线性代数规划中的一个算法，可用来为线性方程组求解。但其算法十分复杂，不常用于加减消元法，求出矩阵的秩，以及求出可逆方阵的逆矩阵

用系数矩阵表示n元一次方程，如
${2x+3y+4z=20x−2y+3z=66x+4y−3z=5⇒(234201−23664−35)\begin{cases} 2x+3y+4z=20\\ x-2y+3z=6\\ 6x+4y-3z=5\\ \end{cases} \ ⇒ \begin{pmatrix} 2&3&4&20\\ 1&-2&3&6\\ 6&4&-3&5\\ \end{pmatrix}$
系数矩阵的运算法则：
• 1.将某一行同时乘以/除以一个非0常数。
• 2.某一行加上或减去另一行的若干倍

演示过程
在这里插入图片描述
$(234201−23664−35)⇒r1∗12(1322101−23664−35)⇒r2−r1,r3−6r1(1322100−721−40−5−15−55)\begin{pmatrix} 2&3&4&20\\ 1&-2&3&6\\ 6&4&-3&5\\ \end{pmatrix} \ ⇒^{r_1*\frac{1}{2}}\begin{pmatrix} 1&\frac{3}{2}&2&10\\ 1&-2&3&6\\ 6&4&-3&5\\ \end{pmatrix}\ ⇒^{r_2-r_1,r_3-6r_1}\begin{pmatrix} 1&\frac{3}{2}&2&10\\ 0&-\frac{7}{2}&1&-4\\ 0&-5&-15&-55\\ \end{pmatrix}$
$⇒r2∗(−27)(13221001−27870−5−15−55)⇒r1−32r2,r3+5r2(1017758701−278700−1157−3457)\ ⇒^{r_2*(-\frac{2}{7})}\begin{pmatrix} 1&\frac{3}{2}&2&10\\ 0&1&-\frac{2}{7}&\frac{8}{7}\\ 0&-5&-15&-55\\ \end{pmatrix}\ ⇒^{r_1-\frac{3}{2}r_2,r_3+5r_2}\begin{pmatrix} 1&0&\frac{17}{7}&\frac{58}{7}\\ 0&1&-\frac{2}{7}&\frac{8}{7}\\ 0&0&-\frac{115}{7}&-\frac{345}{7}\\ \end{pmatrix}$ $⇒r3∗(−7115)(1017758701−27870013)⇒r1−177r3,r2+27r3(100101020013)\ ⇒^{r_3*(-\frac{7}{115})} \begin{pmatrix} 1&0&\frac{17}{7}&\frac{58}{7}\\ 0&1&-\frac{2}{7}&\frac{8}{7}\\ 0&0&1&3\\ \end{pmatrix} \ ⇒^{r_1-\frac{17}{7}r_3,r_2+\frac{2}{7}r_3} \begin{pmatrix} 1&0&0&1\\ 0&1&0&2\\ 0&0&1&3\\ \end{pmatrix}$
即 ${x1=1x2=2x3=3\begin{cases} x_1=1\\ x_2=2\\ x_3=3\\ \end{cases}$

算法流程：
• 从1到n枚举i，第i步定未知数i为主元，从上到下找到第一个未知数i不为0的方程定为主方程（如不存在则跳过这一步）
• 通过乘除系数使得主方程未知数i前面的系数为1 • 对于其他方程，若未知数i前面的系数为k，那么该方程减去主方程的k倍，使得该方程未知数i系数为0
• 时间复杂度O(n^3)

一般来讲，n个未知数n个方程在方程之间线性不相关的情况下，恰好有一组解
如果算法结束后，存在一行全为0，那么被消去的未知数可以任意取值，称为自由元
如果有一行前n个数为0，最后一个数不为0，那么方程无解
当方程组数和未知数数不相等时，也可通过上面方法判定解数
在这里插入图片描述
提一句：多解
解的个数等于自由变量的取值范围的幂
若自由变量的取值范围为 $p$ ,自由变量有 $m$ 个，则解的个数为 $p^m$

线性方程组整数类型解

int a[maxn][maxn];//增广矩阵 
int ans[maxn];
bool Free[maxn];//标记是否为自由变元 int GCD ( int a, int b ) {if ( ! b )return a;return GCD ( b, a % b );
}
int LCM ( int a, int b ) {return a / GCD ( a, b ) * b;
}
int Fabs ( int x ) {if ( x < 0 )return -x;return x;
}int Gauss ( int equ, int var ) {for ( int i = 0;i <= var;i ++ ) {ans[i] = 0;Free[i] = 1;}int row, col, MaxRow;col = 1;for ( row = 1;row <= equ && col < var;row ++, col ++ ) {MaxRow = row;for ( int i = row + 1;i <= equ;i ++ )//寻找当前列绝对值最大的行 if ( Fabs ( a[i][col] ) > Fabs ( a[MaxRow][col] ) )MaxRow = i;if ( MaxRow != row ) {//与第row行交换 for ( int i = row;i <= var;i ++ )swap ( a[row][i], a[MaxRow][i] );}if ( ! a[row][col] ) {//说明col列第row行及以下都是0，就处理当前行的下一列 row --;continue;}for ( int i = row + 1;i <= equ;i ++ ) {//枚举要删去各自col列的变元的系数的行 if ( a[i][col] ) {int lcm = LCM ( Fabs ( a[i][col] ), Fabs ( a[row][col] ) );int T1 = lcm / Fabs ( a[i][col] );int T2 = lcm / Fabs ( a[row][col] );if ( a[i][col] * a[row][col] < 0 )T2 = -T2;for ( int j = col;j <= var;j ++ )a[i][j] = a[i][j] * T1 - a[row][j] * T2;}}}//无解：化简的增广矩阵中存在(0,0,...,a)这种类型 for ( int i = row;i <= equ;i ++ )if ( a[i][col] )return -1;int temp;//无穷解，矩阵中出现(0,0,...0) if ( row < var ) {for ( int i = row - 1;i > 0;i -- ) {// 第i行一定不会是(0, 0, ..., 0)的情况，因为这样的行是在第k行到第equ行.// 同样，第i行一定不会是(0, 0, ..., a), a != 0的情况，这样的无解的int free_num = 0, idx;//用于判断该行中的不确定的变元的个数，如果超过1个，则无法求解，它们仍然为不确定的变元for ( int j = 1;j < var;j ++ )	if ( a[i][j] && Free[j] ) {free_num ++;idx = j;}if ( free_num > 1 ) //无法求解出确定的变元 continue;temp = a[i][var];//说明只有一个不确定的变元，那么就可以解出该变元for ( int j = 1;j < var;j ++ ) {if ( a[i][j] && j != idx )temp -= a[i][j] * ans[j];}ans[idx] = temp / a[i][idx];Free[idx] = 0;}return var - row;//自由变元的个数 }//唯一解的情况，是一个严格的上三角形//1 0 ... 0//0 1 ... 0//0 0 1 . 0 for ( int i = var - 1;i > 0;i -- ) {temp = a[i][var];for ( int j = i + 1;j < var;j ++ )if ( a[i][j] )temp -= a[i][j] * ans[j];//--因为x[i]存的是temp/a[i][i]的值，即是a[i][i]=1时x[i]对应的值
//		if ( temp % a[i][i] )//可以用来判断有浮点数解，无整数解，但这个是整数解模板 
//			return -2;ans[i] = temp / a[i][i];}return 0;
}

线性方程组浮点类型解

#define eps 1e-6
double a[maxn][maxn];
double ans[maxn];
bool Free[maxn];int GCD ( int a, int b ) {if ( ! b )return a;return GCD ( b, a % b );
}
int LCM ( int a, int b ) {return a / GCD ( a, b ) * b;
}int Gauss ( int equ, int var ) {for ( int i = 0;i <= var;i ++ ) {ans[i] = 0;Free[i] = 1;}int row, col, MaxRow;col = 0;for ( row = 0;row < equ && col < var;row ++, col ++ ) {MaxRow = row;for ( int i = row + 1;i < equ;i ++ ) if ( fabs ( a[i][col] ) > fabs ( a[MaxRow][col] ) )MaxRow = i;if ( MaxRow != row ) {for ( int i = row;i <= var;i ++ )swap ( a[row][i], a[MaxRow][i] );}if ( fabs ( a[row][col] ) < eps ) {row --;continue;}for ( int i = row + 1;i < equ;i ++ ) {if ( fabs ( a[i][col] ) > eps ) {double temp = a[i][col] / a[row][col];for ( int j = col;j <= var;j ++ )a[i][j] -= a[row][j] * temp;a[i][col] = 0;}}}for ( int i = row;i < equ;i ++ )if ( fabs ( a[i][col] ) > eps )return -1;double temp;if ( row < var ) {for ( int i = row - 1;i >= 0;i -- ) {int free_num = 0, idx;for ( int j = 0;j < var;j ++ )	if ( a[i][j] && Free[j] ) {free_num ++;idx = j;}if ( free_num > 1 )continue;temp = a[i][var];for ( int j = 0;j < var;j ++ ) {if ( a[i][j] && j != idx )temp -= a[i][j] * ans[j];}ans[idx] = temp / a[i][idx];Free[idx] = 0;}return var - row;}for ( int i = var - 1;i >= 0;i -- ) {temp = a[i][var];for ( int j = i + 1;j < var;j ++ )if ( a[i][j] )temp -= a[i][j] * ans[j];ans[i] = temp / a[i][i];}return 0;
}

模线性方程组

int gcd( int x, int y ) {if( ! y ) return x;else return gcd( y, x % y );
}int inv( int x, int m ) {if( x == 0 ) return 0;if( x == 1 ) return 1;return inv( m % x, m ) * ( m - m / x ) % m;
}int Gauss( int equ, int var ) {int row, col = 1;for( row = 1;row <= equ && col < var;row ++, col ++ ) {int MaxRow = row;for( int i = row + 1;i <= equ;i ++ )if( Fabs( a[i][col] ) > Fabs( a[MaxRow][col] ) )MaxRow = i;if( row != MaxRow ) {for( int i = row;i <= var;i ++ )swap( a[row][i], a[MaxRow][i] );}if( ! a[row][col] ) { row --; continue; }for( int i = row + 1;i <= equ;i ++ ) {if( a[i][col] ) {int d = gcd( Fabs( a[row][col] ), Fabs( a[i][col] ) );int lcm = Fabs( a[row][col] ) / d * Fabs( a[i][col] );int T1 = lcm / Fabs( a[i][col] );int T2 = lcm / Fabs( a[row][col] );if( a[i][col] * a[row][col] < 0 ) T2 = -T2;for( int j = col;j <= var;j ++ ) {a[i][j] = a[i][j] * T1 - a[row][j] * T2;a[i][j] = ( a[i][j] % mod + mod ) % mod;}}}}for( int i = row;i <= equ;i ++ )if( a[i][col] ) return -1;if( row < var ) return var - row;for( int i = var - 1;i;i -- ) {int temp = a[i][var];for( int j = i + 1;j < var;j ++ ) {if( a[i][j] ) {temp -= a[i][j] * x[j];temp = ( temp % mod + mod ) % mod;}}x[i] = temp * inv( a[i][i], mod ) % mod;}return 0;
}

异或方程组

int cnt;
int a[maxn][maxn];
int ans[maxn];
bool Free[maxn];int GCD ( int a, int b ) {if ( ! b )return a;return GCD ( b, a % b );
}
int LCM ( int a, int b ) {return a / GCD ( a, b ) * b;
}int Gauss ( int equ, int var ) {int row, col, MaxRow;col = 0;for ( row = 0;row < equ && col < var;row ++, col ++ ) {MaxRow = row;for ( int i = row + 1;i < equ;i ++ ) if ( fabs ( a[i][col] ) > fabs ( a[MaxRow][col] ) )MaxRow = i;if ( MaxRow != row ) {for ( int i = row;i <= var;i ++ )swap ( a[row][i], a[MaxRow][i] );}if ( a[row][col] == 0 ) {row --;Free[++ cnt] = col;continue;}for ( int i = row + 1;i < equ;i ++ ) {if ( a[i][col] ) {for ( int j = col;j <= var;j ++ )a[i][j] ^= a[row][j];}}}for ( int i = row;i < equ;i ++ )if ( a[i][col] )return -1;if ( row < var ) return var - row;for ( int i = var - 1;i >= 0;i -- ) {ans[i] = a[i][var];for ( int j = i + 1;j < var;j ++ )if ( a[i][j] )ans[i] ^= ( a[i][j] && ans[j] );}return 0;
}

高斯约旦消元

约旦消元

void gauss_jordan ( int equ, int var ) {int row, col = 0;for ( row = 0;row < equ && col < var;row ++, col ++ ) {int MaxRow = row;for ( int i = row + 1;i < equ;i ++ )if ( fabs ( a[i][col] ) > fabs ( a[MaxRow][col] ) )MaxRow = i;if ( MaxRow != row ) {for ( int i = 0;i <= var;i ++ )swap ( a[row][i], a[MaxRow][i] );}if ( fabs ( a[row][col] ) < eps ) {row --;continue;}for ( int i = 0;i < equ;i ++ )if ( i != row )for ( int j = var;j >= row;j -- )a[i][j] -= a[i][col] / a[row][col] * a[row][j];}
}

无解

for ( int i = 0;i < equ;i ++ ) {bool flag = 1;for ( int j = i;j < var;j ++ )if ( fabs ( a[i][j] ) > eps )flag = 0;if ( flag && fabs ( a[i][var] ) > eps )return ! printf ( "-1" );}

无穷解

for ( int i = 0;i < equ;i ++ ) {int free_num = 0;for ( int j = i;j < var;j ++ )if ( fabs ( a[i][j] ) > eps )free_num ++;if ( free_num > 1 )return ! printf ( "0" );}

唯一解

for ( int i = 0;i < equ;i ++ )ans[i] = a[i][var] / a[i][i];for ( int i = 0;i < equ;i ++ )if ( fabs ( ans[i] ) < eps )printf ( "x%d=0\n", i + 1 );elseprintf ( "x%d=%.2f\n", i + 1, ans[i] );