文章目錄

• 理論
• 線性方程組整數類型解
• 線性方程組浮點類型解
• 模線性方程組
• 異或方程組
• 高斯約旦消元
• • 約旦消元
  • 無解
  • 無窮解
  • 唯一解

理論

高斯消元法，是線性代數規劃中的一個算法，可用來為線性方程組求解。但其算法十分復雜，不常用于加減消元法，求出矩陣的秩，以及求出可逆方陣的逆矩陣

用系數矩陣表示n元一次方程，如
$$?\begin{array}{l}2x+3y+4z=20\\ x?2y+3z=6\\ 6x+4y?3z=5\end{array}??\begin{array}{cccc}2& 3& 4& 20\\ 1& ?2& 3& 6\\ 6& 4& ?3& 5\end{array}?$$
系數矩陣的運算法則：
? 1.將某一行同時乘以/除以一個非0常數。
? 2.某一行加上或減去另一行的若干倍

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演示過程

$$?\begin{array}{cccc}2& 3& 4& 20\\ 1& ?2& 3& 6\\ 6& 4& ?3& 5\end{array}?{?}^{r_1?\frac{1}{2}}?\begin{array}{cccc}1& \frac{3}{2}& 2& 10\\ 1& ?2& 3& 6\\ 6& 4& ?3& 5\end{array}?{?}^{r_2?r_1,r_3?6r_1}?\begin{array}{cccc}1& \frac{3}{2}& 2& 10\\ 0& ?\frac{7}{2}& 1& ?4\\ 0& ?5& ?15& ?55\end{array}?$$
$${?}^{r_2?(?\frac{2}{7})}?\begin{array}{cccc}1& \frac{3}{2}& 2& 10\\ 0& 1& ?\frac{2}{7}& \frac{8}{7}\\ 0& ?5& ?15& ?55\end{array}?{?}^{r_1?\frac{3}{2}{r}_2,r_3+5r_2}?\begin{array}{cccc}1& 0& \frac{17}{7}& \frac{58}{7}\\ 0& 1& ?\frac{2}{7}& \frac{8}{7}\\ 0& 0& ?\frac{115}{7}& ?\frac{345}{7}\end{array}?$$$${?}^{r_3?(?\frac{7}{115})}?\begin{array}{cccc}1& 0& \frac{17}{7}& \frac{58}{7}\\ 0& 1& ?\frac{2}{7}& \frac{8}{7}\\ 0& 0& 1& 3\end{array}?{?}^{r_1?\frac{17}{7}{r}_3,r_2+\frac{2}{7}{r}_3}?\begin{array}{cccc}1& 0& 0& 1\\ 0& 1& 0& 2\\ 0& 0& 1& 3\end{array}?$$
即$$?\begin{array}{l}{x}_1=1\\ x_2=2\\ x_3=3\end{array}$$

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算法流程：
? 從1到n枚舉i，第i步定未知數i為主元，從上到下找到第一個未知數i不為0的方程定為主方程（如不存在則跳過這一步）
? 通過乘除系數使得主方程未知數i前面的系數為1 ? 對于其他方程，若未知數i前面的系數為k，那么該方程減去主方程的k倍，使得該方程未知數i系數為0
? 時間復雜度O(n^3)

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一般來講，n個未知數n個方程在方程之間線性不相關的情況下，恰好有一組解
如果算法結束后，存在一行全為0，那么被消去的未知數可以任意取值，稱為自由元
如果有一行前n個數為0，最后一個數不為0，那么方程無解
當方程組數和未知數數不相等時，也可通過上面方法判定解數

提一句：多解
解的個數等于自由變量的取值范圍的冪
若自由變量的取值范圍為$p$ ,自由變量有$m$個，則解的個數為$p^m$

線性方程組整數類型解

int a[maxn][maxn];//增廣矩陣 int ans[maxn]; bool Free[maxn];//標記是否為自由變元 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 ++ )//尋找當前列絕對值最大的行 if ( Fabs ( a[i][col] ) > Fabs ( a[MaxRow][col] ) )MaxRow = i;if ( MaxRow != row ) {//與第row行交換 for ( int i = row;i <= var;i ++ )swap ( a[row][i], a[MaxRow][i] );}if ( ! a[row][col] ) {//說明col列第row行及以下都是0，就處理當前行的下一列 row --;continue;}for ( int i = row + 1;i <= equ;i ++ ) {//枚舉要刪去各自col列的變元的系數的行 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;}}}//無解：化簡的增廣矩陣中存在(0,0,...,a)這種類型 for ( int i = row;i <= equ;i ++ )if ( a[i][col] )return -1;int temp;//無窮解，矩陣中出現(0,0,...0) if ( row < var ) {for ( int i = row - 1;i > 0;i -- ) {// 第i行一定不會是(0, 0, ..., 0)的情況，因為這樣的行是在第k行到第equ行.// 同樣，第i行一定不會是(0, 0, ..., a), a != 0的情況，這樣的無解的int free_num = 0, idx;//用于判斷該行中的不確定的變元的個數，如果超過1個，則無法求解，它們仍然為不確定的變元for ( int j = 1;j < var;j ++ ) if ( a[i][j] && Free[j] ) {free_num ++;idx = j;}if ( free_num > 1 ) //無法求解出確定的變元 continue;temp = a[i][var];//說明只有一個不確定的變元，那么就可以解出該變元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;//自由變元的個數 }//唯一解的情況，是一個嚴格的上三角形//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];//--因為x[i]存的是temp/a[i][i]的值，即是a[i][i]=1時x[i]對應的值 // if ( temp % a[i][i] )//可以用來判斷有浮點數解，無整數解，但這個是整數解模板 // return -2;ans[i] = temp / a[i][i];}return 0; }

線性方程組浮點類型解

#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; }

模線性方程組

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; }

異或方程組

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; }

高斯約旦消元

約旦消元

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];} }

無解

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" );}

無窮解

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] );

溫馨提示，不保證code的絕對正確，反正我是過了
錯了別打我，至少別打臉（跑ε=ε=ε=(_￣▽￣)

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