拉格朗日插值法

拉格朗日插值法

   
     

    
      
    
      #include
    
      "
    
      iostream
    
      "
    
      
#include
    
      "
    
      math.h
    
      "
    
      

    
      using
    
       
    
      namespace
    
       std;

    
      int
    
       n;

    
      int
    
       main()
{
 
    
      int
    
       N;
 
    
      double
    
       x[
    
      3
    
      ]
    
      =
    
      {
    
      0.2
    
      ,
    
      0.3
    
      ,
    
      0.4
    
      }, y[
    
      3
    
      ]
    
      =
    
      {
    
      1.2214
    
      ,
    
      1.3499
    
      ,
    
      1.4918
    
      }, Y
    
      =
    
      0
    
      , X1
    
      =
    
      0.285
    
      ,b,m;
 
    
      int
    
       i,j,k;
 
    
      for
    
      (k
    
      =
    
      0
    
      ;k
    
      <
    
      3
    
      ;k
    
      ++
    
      )
 {
 
    
      double
    
       p
    
      =
    
      1
    
      ;
 
    
      for
    
      (j
    
      =
    
      0
    
      ;j
    
      <
    
      3
    
      ;j
    
      ++
    
      )
 {
 
    
      if
    
      (j
    
      ==
    
      k) 
    
      continue
    
      ;
 
    
      else
    
       
 {
 p
    
      =
    
      p
    
      *
    
      (X1
    
      -
    
      x[j])
    
      /
    
      (x[k]
    
      -
    
      x[j]);
 }
 
 }
 Y
    
      =
    
      Y
    
      +
    
      p
    
      *
    
      y[k];
 }
 
    
      /*
    
      X1=0.285时的近似值
    
      */
    
      
 printf(
    
      "
    
      e(%.3lf)=%.6lf\n
    
      "
    
      ,X1,Y);
 cout
    
      <<
    
      "
    
      Please Input N value( 2 or 3 )
    
      "
    
      <<
    
      endl
    
      <<
    
      "
    
      N=
    
      "
    
      ;
 
    
      while
    
      (cin
    
      >>
    
      N)
 {
 
 
    
      if
    
      (N
    
      ==
    
      2
    
      ||
    
      N
    
      ==
    
      3
    
      )
 {
 
    
      double
    
       x[
    
      3
    
      ]
    
      =
    
      {
    
      0.2
    
      ,
    
      0.3
    
      ,
    
      0.4
    
      }, y[
    
      3
    
      ]
    
      =
    
      {
    
      1.2214
    
      ,
    
      1.3499
    
      ,
    
      1.4918
    
      }, Y
    
      =
    
      0
    
      , X
    
      =
    
      0.285
    
      ;
 
    
      for
    
      (k
    
      =
    
      0
    
      ;k
    
      <
    
      N;k
    
      ++
    
      )
 {
 
    
      double
    
       p
    
      =
    
      1
    
      ;
 
    
      for
    
      (j
    
      =
    
      0
    
      ;j
    
      <
    
      N;j
    
      ++
    
      )
 { 

 
    
      if
    
      (j
    
      ==
    
      k) 
    
      continue
    
      ;
 
    
      else
    
       
 p
    
      =
    
      p
    
      *
    
      (X
    
      -
    
      x[j])
    
      /
    
      (x[k]
    
      -
    
      x[j]);
 }
 Y
    
      =
    
      Y
    
      +
    
      p
    
      *
    
      y[k];
 }
 printf(
    
      "
    
      L(%d)=%.6lf\n
    
      "
    
      , N
    
      -
    
      1
    
      ,Y);
 
    
      double
    
       X2;Y
    
      =
    
      0
    
      ;

 m
    
      =
    
      1
    
      ;
 
    
      if
    
      (N
    
      ==
    
      2
    
      ) X2
    
      =
    
      0.3
    
      ;
 
    
      else
    
       X2
    
      =
    
      0.4
    
      ;
 
    
      for
    
      (k
    
      =
    
      0
    
      ;k
    
      <
    
      N;k
    
      ++
    
      )
 {
 
    
      double
    
       p
    
      =
    
      1
    
      ;
 
    
      for
    
      (j
    
      =
    
      0
    
      ;j
    
      <
    
      N;j
    
      ++
    
      )
 {
 
    
      if
    
      (j
    
      ==
    
      k) 
    
      continue
    
      ;
 
    
      else
    
       
 {
 p
    
      =
    
      p
    
      *
    
      (X2
    
      -
    
      x[j])
    
      /
    
      (x[k]
    
      -
    
      x[j]);
 }
 
 }
 Y
    
      =
    
      Y
    
      +
    
      p
    
      *
    
      y[k];
 m
    
      *=
    
      (X1
    
      -
    
      x[k]);
 }
 
    
      /*
    
      N==2或N==3的截断误差
    
      */
    
      
 printf(
    
      "
    
      R%d(%.6lf)<%.5lf\n
    
      "
    
      ,N
    
      -
    
      1
    
      ,X1,fabs(Y
    
      *
    
      m
    
      /
    
      6
    
      ));
 
    
      //
    
       cout<<"Please Input N value(2 or 3)"<<endl<<"N=";
    
      

    
       }
 
    
      else
    
      
 {
 cout
    
      <<
    
      "
    
      error!
    
      "
    
      <<
    
      endl;
 }
 cout
    
      <<
    
      "
    
      Please Input N value( 2 or 3 )
    
      "
    
      <<
    
      endl
    
      <<
    
      "
    
      N=
    
      "
    
      ;
 } 
}
   
     

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