Solving a system of linear equations by Gaussian elimination(row reduction plus Matrix Rotation skill) 聯立方程式之高斯消去法併用矩陣旋轉技巧

  1 //++++++++++++++++++++++++++
  2  //      SystemEq.cpp
  3  //++++++++++++++++++++++++++
  4  
  5  ///#include header
  6  //#include<conio.h>
  7  #include<bits/stdc++.h>
  8  
  9  ///main_in
 10  int main(int argc,char *argv[])
 11  {
 12      ///declare
 13      int i, j, iEqs;
 14      float fCof[20][20], x[20];
 15      float dfCof[20][40];
 16      
 17      ///system cls
 18      system("cls");
 19      
 20      ///Prompt input_eqs
 21      printf("Enter the number of equations : ");
 22      
 23      ///Keyin _eqs
 24      scanf("%d", &iEqs);
 25      
 26      ///Prompt input_cof
 27      printf("\nEnter the coefficients of the equations :\n\n");
 28      
 29      ///Xru 0 to colmax-1
 30      for(i = 0 ; i < iEqs ; i++)
 31      {
 32          ///Xru 0 to rowmax-1
 33          for(j = 0 ; j < iEqs ; j++)
 34          {
 35              ///Prompt input_Xij
 36              printf("X%d%d = ", i + 1, j + 1);
 37              
 38              ///Keyin COEFFICIENT-i.j
 39              scanf("%f", &dfCof[i][j]);
 40              dfCof[i][j+iEqs]=dfCof[i][j];
 41              
 42              ///+
 43          }//end for
 44          
 45          ///Prompt input_Yij
 46          printf("Y%d = ", i + 1);
 47          
 48          ///Keyin COEFFICIENT rowmax
 49          scanf("%f", &dfCof[i][iEqs*2]);
 50          
 51          ///+
 52      }//end for
 53      
 54      ///matrix left_shift
 55      int iShift;
 56      for(iShift = iEqs ; iShift > 0  ; iShift--)
 57      {
 58          ///Copy Block dfCof2fCof
 59          for(i = 0 ; i < iEqs ; i++) for(j = 0 ; j < iEqs ; j++) fCof[i][j] = dfCof[i][j+iShift];
 60          
 61          ///Copy Y_col
 62          for(i = 0 ; i < iEqs ; i++) fCof[i][iEqs] = dfCof[i][iEqs*2];
 63          
 64          ///Reduce Cycle iRdsCyc=0,1 (0~iEQs-1)
 65          int iRdsCyc;
 66          for(iRdsCyc = 0 ; iRdsCyc < iEqs - 1 ; iRdsCyc++)
 67          {
 68              ///Renew row count iRenuRowCnt
 69              int iRenuRowCnt;
 70              for(iRenuRowCnt = iRdsCyc + 1 ; iRenuRowCnt < iEqs ; iRenuRowCnt++)
 71              {
 72                  ///XpnOvrXnn
 73                  double XpnOvrXnn;
 74                  XpnOvrXnn = fCof[iRenuRowCnt][iRdsCyc] / fCof[iRdsCyc][iRdsCyc];
 75                  
 76                  ///Renew row iRenuRow=4,3 =iRdsCyc~iEQs
 77                  int iRenuRow;
 78                  for(iRenuRow = iRdsCyc ; iRenuRow < iEqs + 1 ; iRenuRow++)
 79                  {
 80                      fCof[iRenuRowCnt][iRenuRow] = fCof[iRenuRowCnt][iRenuRow] - fCof[iRdsCyc][iRenuRow] *XpnOvrXnn ;
 81                  }
 82                  
 83                  ///+
 84              }//end for
 85              
 86              ///+
 87          }//end for Reduce Cycle
 88          
 89          ///Xn=x[EQs-1] =Y/X
 90          x[iShift-1] = fCof[iEqs-1][iEqs] / fCof[iEqs-1][iEqs-1];
 91          
 92          ///+
 93      }//end for
 94      
 95      ///Prompt result
 96      printf("\nThe result is :\n");
 97      
 98      ///Display result
 99      for(i = 0 ; i < iEqs ; i++) {printf("\nsolution of x%d = %.2f", i + 1, x[i]);}
100      
101      ///system pause
102      printf("\n\n");
103      system("pause");
104      
105      ///main_end
106      return 0;
107  } //main_end
108  

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