高斯消元 Java 高精度版 HDU 2449 Gauss Elimination

http://acm.hdu.edu.cn/showproblem.php?pid=2449

题意 ：

 纯高斯消元 ；

输入  n 行 ，每行  n+1个数带代表 系数和 值  

ai1,ai2,ai3…..ain, bi

ai1*x1+ai2*x2+......ain*xn=bi

求解  xi  若没有整数解 输出 分数 ，

若没有解 输出  

No solution.

 

copy 别人的 高精度 Gauss （留着用）（不会java 啊 ）

  1 import java.util.*;
  2 import java.math.*;
  3 
  4 class fraction{
  5     BigInteger a, b;
  6     public fraction(){ 
  7         a = new BigInteger("0");
  8         b = new BigInteger("1");
  9     }
 10     
 11     fraction( BigInteger a0, BigInteger b0){
 12         this.a = a0; this.b = b0;
 13     }
 14     void reduction(){
 15         BigInteger tmp = a.gcd( b );
 16         a = a.divide( tmp );
 17         b = b.divide( tmp );
 18         if( b.compareTo( BigInteger.ZERO ) == - 1 )
 19         {
 20             b = b.multiply( BigInteger.valueOf( -1 ));
 21             a = a.multiply( BigInteger.valueOf( -1 ));
 22         }
 23     }
 24     fraction add( fraction t ){
 25         fraction tmp = new fraction( a.multiply( t.b ).add( b.multiply( t.a )) , b.multiply(t.b) );
 26         tmp.reduction();
 27         return tmp;
 28     }
 29     fraction sub( fraction t ){
 30         fraction tmp = new fraction( a.multiply( t.b ).subtract( b.multiply( t.a )) , b.multiply(t.b) );
 31         tmp.reduction();
 32         return tmp;
 33     }
 34     fraction mult( fraction t){
 35         fraction tmp = new fraction( a.multiply( t.a ), b.multiply( t.b ));
 36         tmp.reduction();
 37         return tmp;
 38     }
 39     fraction div( fraction t){
 40         fraction tmp = new fraction( a.multiply( t.b ), b.multiply( t.a ));
 41         tmp.reduction();
 42         return tmp;
 43     }
 44     public void abs(){
 45         if( this.a.compareTo( BigInteger.ZERO ) == - 1 ){
 46             this.a = this.a.multiply( BigInteger.valueOf( -1 ));
 47         }
 48     }
 49     void out(){
 50         this.reduction();
 51         if( b.compareTo( BigInteger.ONE ) == 0 )
 52             System.out.println(a);
 53         else
 54             System.out.println(a+"/"+b);
 55     }
 56     
 57     boolean biger( fraction p ){
 58         fraction tmp = new fraction ( a, b );
 59         fraction t = new fraction(p.a,p.b);
 60         //t = p;
 61         tmp.reduction();
 62         if( tmp.a.compareTo( BigInteger.ZERO ) == - 1 ){
 63             tmp.a = tmp.a.multiply( BigInteger.valueOf( -1 ));
 64         }
 65         if( t.a.compareTo( BigInteger.ZERO ) == - 1 ){
 66             t.a = t.a.multiply( BigInteger.valueOf( -1 ));
 67         }
 68         tmp = tmp.sub( t );
 69         return tmp.a.compareTo( BigInteger.ZERO ) > -1;
 70     }
 71     
 72 }
 73 
 74 public class Main{
 75 
 76     public static void lup_solve( fraction x[],fraction y[], fraction L[][], fraction U[][],int pi[],fraction b[],int n)
 77     {
 78         int i, j;
 79         fraction z = new fraction( BigInteger.ZERO , BigInteger.ONE);
 80         fraction sum = z;//double sum;
 81         for ( i = 0 ; i < n ; i ++ ){
 82             sum = z; //sum = 0;
 83             for ( j = 0 ; j < i ; j ++ ){
 84                 sum = sum.add( L[i][j].mult( y[j] ));//sum += L[i][j] * y[ j ];
 85             }
 86             y[i] = b[ pi[i] ].sub( sum );//y[i] = b[ pi[i] ] - sum;
 87         }
 88         for ( i = n - 1 ; i >= 0 ; i -- ){
 89             sum = z ; //sum = 0;
 90             for ( j = i + 1 ; j < n ; j ++ ){
 91                 sum = sum.add( U[i][j].mult( x[j] ));//sum += U[i][j] * x[ j ];
 92             }
 93             x[i] = (y[i].sub( sum )).div( U[i][i] );//x[i] = (y[i] - sum)/U[i][i];
 94         }
 95     }
 96     
 97     public static int lup_decomposition( fraction a[][] , int n , int pi[] )
 98     {
 99         int i, j, k, k1 = 0 ;
100         fraction p = new fraction(BigInteger.valueOf(0), BigInteger.ONE ), z = new fraction( BigInteger.valueOf(0), BigInteger.ONE );
101         for ( i = 0 ; i < n ; i ++ )
102             pi[i] = i;// 置换
103             
104         for ( k = 0 ; k < n ; k ++ ){
105             p = z;
106             for ( i = k ; i < n ; i ++ )
107             {
108                 if(  a[i][k].biger( p ) )
109                 {
110                     p = new fraction( a[i][k].a,a[i][k].b) ;
111                     k1 = i;
112                 }
113             }
114             if( p.a.compareTo( BigInteger.ZERO ) == 0 ){
115                 return 0 ;// error
116             }
117             fraction tmp;
118             
119             int t = pi[ k ]; pi[ k ] = pi[ k1 ]; pi[k1] = t;
120             for ( i = 0 ; i < n ; i ++ ){
121                 tmp = a[ k ][i];  a[ k ][i] = a[ k1 ][i];  a[k1][i] = tmp;
122             }//swap( a[k][i], a[k1][i] );
123             for ( i = k + 1 ; i < n ; i ++ )
124             {
125                 a[i][k] = a[i][k].div( a[k][k] );
126                 for ( j = k + 1 ; j < n ; j ++ )
127                     a[i][j] = a[i][j].sub( a[i][k].mult(a[k][j]));// - a[i][k] * a[k][j] ;
128             }
129         }
130         return 1;
131     }
132     
133     public static void check(fraction a[][], fraction x[], int n){
134         int i, j;
135         fraction sum, z = new fraction( BigInteger.ZERO , BigInteger.ONE);
136         for ( i = 0 ;  i < n ; i++ ){
137             sum = z;
138             for ( j = 0 ;j < n ; j ++ )
139             {
140                 sum = sum.add( a[i][j].mult( x[j] ));
141             }
142             sum.out();
143         }
144     }
145     
146     public static void main(String[] agrs){
147         Scanner cin = new Scanner( System.in );
148         int i, j;
149         int n;
150         while( cin.hasNextInt() )
151         {
152             //任何函数都要和一个class相连
153             n = cin.nextInt();
154             int pi[] = new int[n];
155             fraction a[][] = new fraction[n][n];
156             fraction aa[][] = new fraction[n][n];
157             fraction B[] = new fraction[n];
158             fraction x[] = new fraction[n];
159             fraction y[] = new fraction[n];
160             
161             for ( i = 0 ; i < n ; i ++ )
162             {
163                 for ( j = 0 ;j < n ; j ++ ){
164                     a[i][j] = new fraction( BigInteger.valueOf(0),BigInteger.valueOf(1));
165                     a[i][j].a = cin.nextBigInteger();
166                     aa[i][j] = new fraction( BigInteger.valueOf(0),BigInteger.valueOf(1));
167                     aa[i][j] = a[i][j];
168                 }
169                 B[i] = new fraction( BigInteger.valueOf(0),BigInteger.valueOf(1));
170                 B[i].a = cin.nextBigInteger();
171                 x[i] = new fraction( BigInteger.valueOf(0),BigInteger.valueOf(1));
172                 y[i] = new fraction( BigInteger.valueOf(0),BigInteger.valueOf(1));
173             }
174             if( 1 == lup_decomposition( a, n, pi) )
175             {
176                 lup_solve( x, y, a, a, pi, B, n);
177                 for ( i = 0 ;i < n; i ++)
178                     x[i].out();
179                 //check( aa, x, n);
180             }
181             else
182             {
183                 System.out.println("No solution.");
184             }
185             System.out.println();
186         }
187     }
188 }