python n元一次方程模糊解
#include <math.h>
#include <conio.h>
#include<stdio.h>
void Iteration_f(double);
int Epsilon_f (double*,double*,int,double);
void matrix_f(void);
void Initialize_f(void);
void Rearrange_f(void);
void Solve_f(double);
void ShowAnswer(void);
void Output_f(void);
void Input_f(void);
int nRow,nColumn;
double **dlpMat;
int nVarNum;
double *dlpVar;
int nIterCount;
Matrix_f(void)
python printf输出格式{
  int nCounti;
  /*creates my matrix dynamically, but there are still no values in the positions */
  dlpMat=(double **) malloc(sizeof( double) *nRow*nColumn);
}
void Initialize_f (void)
{
  printf("How many ROWS?");
  scanf("%d", &nRow);
  printf("How many COLUMNS?");
  scanf("%d", &nColumn);
}
/**********************************
*      高斯-赛德尔矩阵变换函数    *
*      rearrange function        *
**********************************/
void Rearrange_f(void)
{
  int nCounti, nCountj;
  double dlCoeff;
  nVarNum=nColumn-1;
  /* 'variable' will contain the solution set */
  /* they will get initialized with the first guess in the 'solve' function */
  dlpVar=(double*)malloc(sizeof(double) * nVarNum);
  /*divides all terms by the desired variables    (the dlpVar which is being solved for) coefficient*/
  for(nCounti=0;nCounti<nRow;nCounti++)
  {
      dlCoeff=dlpMat[nCounti][nCounti];
      for(nCountj=0;nCountj<nColumn;nCountj++)
      {
    dlpMat[nCounti][nCountj]/=dlCoeff;
      }
  }
  /* all variables except the diagonal are being brought to the other side */
  for( nCounti=0;nCounti<nRow;nCounti++)
  {
      for(nCountj=0;nCountj<nColumn-1;nCountj++)
      {
    dlpMat[nCounti][nCountj]*=-1;
      }
      dlpMat[nCounti][nCounti]=0;
  }
}
/***********************************
*      iteration funtion--迭代函数    *
***********************************/
void Iteration_f(double dlLambda)
{
  int nCounti, nCountj;
  /* 'last' is for the relaxation equation */
  double dlLast;
  for(nCounti=0;nCounti<nVarNum;nCounti++)
  {
      dlLast=dlpVar[nCounti];
      dlpVar[nCounti]=0;
      for( nCountj=0;nCountj<nVarNum;nCountj++)
      {
      dlpVar[nCounti]+=dlpMat[nCounti][nCountj]*dlpVar[nCountj];
      }
      dlpVar[nCounti]+=dlpMat[nCounti][nColumn-1];
      /* new value after relaxation */
      dlpVar[nCounti]=dlLast+dlLambda *(dlpVar[nCounti]-dlLast);
  }
}
/*********************************
*        solve function--求解函数    *
*********************************/
void Solve_f(double dlLambda)
{
  double dlCriterion;
  int nCounti, nCountj;
  double *dlpNew, *dlpLast;
  for (nCounti=0;nCounti<nVarNum;nCounti++)

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