10个重要的算法C语言实现源代码：拉格朗日，牛顿插值，高斯，龙贝格，牛顿迭代，牛顿-科特斯，雅克比，秦九昭，幂法，高斯塞德尔

 

 

6. 牛顿-科特斯求积公式，求定积分

C/C++ code

    
    
    
    

     
     
     
     
     #include
     
     
     
     <
     
     
     
     stdio.h
     
     
     
     >
     
     
     
     #include
     
     
     
     <
     
     
     
     math.h
     
     
     
     >
     
     
     
     
     
     
     
     int
     
     
     
      NC(a,h,n,r,f)
     
     
     
     float
     
     
     
      (
     
     
     
     *
     
     
     
     a)[];
     
     
     
     float
     
     
     
      h;
     
     
     
     int
     
     
     
      n,f;
     
     
     
     float
     
     
     
      
     
     
     
     *
     
     
     
     r;{ 
     
     
     
     int
     
     
     
      nn,i; 
     
     
     
     float
     
     
     
      ds; 
     
     
     
     if
     
     
     
     (n
     
     
     
     >
     
     
     
     1000
     
     
     
     ||
     
     
     
     n
     
     
     
     <
     
     
     
     2
     
     
     
     ) { 
     
     
     
     if
     
     
     
      (f) printf(
     
     
     
     "
     
     
     
     /n Faild! Check if 1<n<1000!/n
     
     
     
     "
     
     
     
     ,n); 
     
     
     
     return
     
     
     
     (
     
     
     
     -
     
     
     
     1
     
     
     
     );}
     
     
     
     if
     
     
     
     (n
     
     
     
     ==
     
     
     
     2
     
     
     
     ){ 
     
     
     
     *
     
     
     
     r
     
     
     
     =
     
     
     
     0.5
     
     
     
     *
     
     
     
     ((
     
     
     
     *
     
     
     
     a)[
     
     
     
     0
     
     
     
     ]
     
     
     
     +
     
     
     
     (
     
     
     
     *
     
     
     
     a)[
     
     
     
     1
     
     
     
     ])
     
     
     
     *
     
     
     
     (h);
     
     
     
     return
     
     
     
     (
     
     
     
     0
     
     
     
     );}
     
     
     
     if
     
     
     
      (n
     
     
     
     -
     
     
     
     4
     
     
     
     ==
     
     
     
     0
     
     
     
     ) { 
     
     
     
     *
     
     
     
     r
     
     
     
     =
     
     
     
     0
     
     
     
     ;
     
     
     
     *
     
     
     
     r
     
     
     
     =*
     
     
     
     r
     
     
     
     +
     
     
     
     0.375
     
     
     
     *
     
     
     
     (h)
     
     
     
     *
     
     
     
     ((
     
     
     
     *
     
     
     
     a)[n
     
     
     
     -
     
     
     
     4
     
     
     
     ]
     
     
     
     +
     
     
     
     3
     
     
     
     *
     
     
     
     (
     
     
     
     *
     
     
     
     a)[n
     
     
     
     -
     
     
     
     3
     
     
     
     ]
     
     
     
     +
     
     
     
     3
     
     
     
     *
     
     
     
     (
     
     
     
     *
     
     
     
     a)[n
     
     
     
     -
     
     
     
     2
     
     
     
     ]
     
     
     
     +
     
     
     
     (
     
     
     
     *
     
     
     
     a)[n
     
     
     
     -
     
     
     
     1
     
     
     
     ]);
     
     
     
     return
     
     
     
     (
     
     
     
     0
     
     
     
     );}
     
     
     
     if
     
     
     
     (n
     
     
     
     /
     
     
     
     2
     
     
     
     -
     
     
     
     (n
     
     
     
     -
     
     
     
     1
     
     
     
     )
     
     
     
     /
     
     
     
     2
     
     
     
     <=
     
     
     
     0
     
     
     
     )nn
     
     
     
     =
     
     
     
     n;
     
     
     
     else
     
     
     
     nn
     
     
     
     =
     
     
     
     n
     
     
     
     -
     
     
     
     3
     
     
     
     ;ds
     
     
     
     =
     
     
     
     (
     
     
     
     *
     
     
     
     a)[
     
     
     
     0
     
     
     
     ]
     
     
     
     -
     
     
     
     (
     
     
     
     *
     
     
     
     a)[nn
     
     
     
     -
     
     
     
     1
     
     
     
     ];
     
     
     
     for
     
     
     
     (i
     
     
     
     =
     
     
     
     2
     
     
     
     ;i
     
     
     
     <=
     
     
     
     nn;i
     
     
     
     =
     
     
     
     i
     
     
     
     +
     
     
     
     2
     
     
     
     )ds
     
     
     
     =
     
     
     
     ds
     
     
     
     +
     
     
     
     4
     
     
     
     *
     
     
     
     (
     
     
     
     *
     
     
     
     a)[i
     
     
     
     -
     
     
     
     1
     
     
     
     ]
     
     
     
     +
     
     
     
     2
     
     
     
     *
     
     
     
     (
     
     
     
     *
     
     
     
     a)[i];
     
     
     
     *
     
     
     
     r
     
     
     
     =
     
     
     
     ds
     
     
     
     *
     
     
     
     (h)
     
     
     
     /
     
     
     
     3
     
     
     
     ;
     
     
     
     if
     
     
     
     (n
     
     
     
     >
     
     
     
     nn)
     
     
     
     *
     
     
     
     r
     
     
     
     =*
     
     
     
     r
     
     
     
     +
     
     
     
     0.375
     
     
     
     *
     
     
     
     (h)
     
     
     
     *
     
     
     
     ((
     
     
     
     *
     
     
     
     a)[n
     
     
     
     -
     
     
     
     4
     
     
     
     ]
     
     
     
     +
     
     
     
     3
     
     
     
     *
     
     
     
     (
     
     
     
     *
     
     
     
     a)[n
     
     
     
     -
     
     
     
     3
     
     
     
     ]
     
     
     
     +
     
     
     
     3
     
     
     
     *
     
     
     
     (
     
     
     
     *
     
     
     
     a)[n
     
     
     
     -
     
     
     
     2
     
     
     
     ]
     
     
     
     +
     
     
     
     (
     
     
     
     *
     
     
     
     a)[n
     
     
     
     -
     
     
     
     1
     
     
     
     ]);
     
     
     
     return
     
     
     
     (
     
     
     
     0
     
     
     
     );}main(){ 
     
     
     
     float
     
     
     
      h,r; 
     
     
     
     int
     
     
     
      n,ntf,f; 
     
     
     
     int
     
     
     
      i; 
     
     
     
     float
     
     
     
      a[
     
     
     
     16
     
     
     
     ];printf(
     
     
     
     "
     
     
     
     Input the x[i](16):/n
     
     
     
     "
     
     
     
     ); 
     
     
     
     for
     
     
     
     (i
     
     
     
     =
     
     
     
     0
     
     
     
     ;i
     
     
     
     <=
     
     
     
     15
     
     
     
     ;i
     
     
     
     ++
     
     
     
     ) scanf(
     
     
     
     "
     
     
     
     %d
     
     
     
     "
     
     
     
     ,
     
     
     
     &
     
     
     
     a[i]);h
     
     
     
     =
     
     
     
     0.2
     
     
     
     ;f
     
     
     
     =
     
     
     
     0
     
     
     
     ;ntf
     
     
     
     =
     
     
     
     NC(a,h,n,
     
     
     
     &
     
     
     
     r,f);
     
     
     
     if
     
     
     
     (ntf
     
     
     
     ==
     
     
     
     0
     
     
     
     ) printf(
     
     
     
     "
     
     
     
     /nR=%f/n
     
     
     
     "
     
     
     
     ,r); 
     
     
     
     else
     
     
     
      printf(
     
     
     
     "
     
     
     
     /n Wrong!Return code=%d/n
     
     
     
     "
     
     
     
     ,ntf); getch();}
    
    
    
    

7.雅克比迭代，求解方程近似解

C/C++ code

    
    
    
    

     
     
     
     
     #include 
     
     
     
     <
     
     
     
     stdio.h
     
     
     
     >
     
     
     
     #include 
     
     
     
     <
     
     
     
     math.h
     
     
     
     >
     
     
     
     
     
     
     
     #define
     
     
     
      N 20
     
     
     
     
     
     
     
     #define
     
     
     
      MAX 100
     
     
     
     
     
     
     
     #define
     
     
     
      e 0.00001
     
     
     
     
     
     
     
     int
     
     
     
      main(){ 
     
     
     
     int
     
     
     
      n; 
     
     
     
     int
     
     
     
      i,j,k; 
     
     
     
     float
     
     
     
      t; 
     
     
     
     float
     
     
     
      a[N][N],b[N][N],c[N],g[N],x[N],h[N]; printf(
     
     
     
     "
     
     
     
     /nInput dim of n:
     
     
     
     "
     
     
     
     ); scanf(
     
     
     
     "
     
     
     
     %d
     
     
     
     "
     
     
     
     ,
     
     
     
     &
     
     
     
     n); 
     
     
     
     if
     
     
     
     (n
     
     
     
     >
     
     
     
     N) { printf(
     
     
     
     "
     
     
     
     Faild! Check if 0<n<N!/n
     
     
     
     "
     
     
     
     ); getch(); 
     
     
     
     return
     
     
     
      
     
     
     
     1
     
     
     
     ; } 
     
     
     
     if
     
     
     
     (n
     
     
     
     <=
     
     
     
     0
     
     
     
     ) {printf(
     
     
     
     "
     
     
     
     Faild! Check if 0<n<N!/n
     
     
     
     "
     
     
     
     ); getch(); 
     
     
     
     return
     
     
     
      
     
     
     
     1
     
     
     
     ;} printf(
     
     
     
     "
     
     
     
     Input a[i,j],i,j=0…%d:/n
     
     
     
     "
     
     
     
     ,n
     
     
     
     -
     
     
     
     1
     
     
     
     ); 
     
     
     
     for
     
     
     
     (i
     
     
     
     =
     
     
     
     0
     
     
     
     ;i
     
     
     
     <
     
     
     
     n;i
     
     
     
     ++
     
     
     
     ) 
     
     
     
     for
     
     
     
     (j
     
     
     
     =
     
     
     
     0
     
     
     
     ;j
     
     
     
     <
     
     
     
     n;j
     
     
     
     ++
     
     
     
     ) scanf(
     
     
     
     "
     
     
     
     %f
     
     
     
     "
     
     
     
     ,
     
     
     
     &
     
     
     
     a[i][j]); printf(
     
     
     
     "
     
     
     
     Input c[i],i=0…%d:/n
     
     
     
     "
     
     
     
     ,n
     
     
     
     -
     
     
     
     1
     
     
     
     ); 
     
     
     
     for
     
     
     
     (i
     
     
     
     =
     
     
     
     0
     
     
     
     ;i
     
     
     
     <
     
     
     
     n;i
     
     
     
     ++
     
     
     
     )scanf(
     
     
     
     "
     
     
     
     %f
     
     
     
     "
     
     
     
     ,
     
     
     
     &
     
     
     
     c[i]); 
     
     
     
     for
     
     
     
     (i
     
     
     
     =
     
     
     
     0
     
     
     
     ;i
     
     
     
     <
     
     
     
     n;i
     
     
     
     ++
     
     
     
     ) 
     
     
     
     for
     
     
     
     (j
     
     
     
     =
     
     
     
     0
     
     
     
     ;j
     
     
     
     <
     
     
     
     n;j
     
     
     
     ++
     
     
     
     ) { b[i][j]
     
     
     
     =-
     
     
     
     a[i][j]
     
     
     
     /
     
     
     
     a[i][i]; g[i]
     
     
     
     =
     
     
     
     c[i]
     
     
     
     /
     
     
     
     a[i][i]; } 
     
     
     
     for
     
     
     
     (i
     
     
     
     =
     
     
     
     0
     
     
     
     ;i
     
     
     
     <
     
     
     
     MAX;i
     
     
     
     ++
     
     
     
     ) { 
     
     
     
     for
     
     
     
     (j
     
     
     
     =
     
     
     
     0
     
     
     
     ;j
     
     
     
     <
     
     
     
     n;j
     
     
     
     ++
     
     
     
     ) h[j]
     
     
     
     =
     
     
     
     g[j]; { 
     
     
     
     for
     
     
     
     (k
     
     
     
     =
     
     
     
     0
     
     
     
     ;k
     
     
     
     <
     
     
     
     n;k
     
     
     
     ++
     
     
     
     ) { 
     
     
     
     if
     
     
     
     (j
     
     
     
     ==
     
     
     
     k) 
     
     
     
     continue
     
     
     
     ; h[j]
     
     
     
     +=
     
     
     
     b[j][k]
     
     
     
     *
     
     
     
     x[k]; } } t
     
     
     
     =
     
     
     
     0
     
     
     
     ; 
     
     
     
     for
     
     
     
     (j
     
     
     
     =
     
     
     
     0
     
     
     
     ;j
     
     
     
     <
     
     
     
     n;j
     
     
     
     ++
     
     
     
     ) 
     
     
     
     if
     
     
     
     (t
     
     
     
     <
     
     
     
     fabs(h[j]
     
     
     
     -
     
     
     
     x[j])) t
     
     
     
     =
     
     
     
     fabs(h[j]
     
     
     
     -
     
     
     
     x[j]); 
     
     
     
     for
     
     
     
     (j
     
     
     
     =
     
     
     
     0
     
     
     
     ;j
     
     
     
     <
     
     
     
     n;j
     
     
     
     ++
     
     
     
     ) x[j]
     
     
     
     =
     
     
     
     h[j]; 
     
     
     
     if
     
     
     
     (t
     
     
     
     <
     
     
     
     e) { printf(
     
     
     
     "
     
     
     
     x_i=/n
     
     
     
     "
     
     
     
     ); 
     
     
     
     for
     
     
     
     (i
     
     
     
     =
     
     
     
     0
     
     
     
     ;i
     
     
     
     <
     
     
     
     n;i
     
     
     
     ++
     
     
     
     ) printf(
     
     
     
     "
     
     
     
     x[%d]=%f/n
     
     
     
     "
     
     
     
     ,i,x[i]); getch(); 
     
     
     
     return
     
     
     
      
     
     
     
     0
     
     
     
     ; } printf(
     
     
     
     "
     
     
     
     after %d repeat , return/n
     
     
     
     "
     
     
     
     ,MAX); getch(); 
     
     
     
     return
     
     
     
      
     
     
     
     1
     
     
     
     ; } getch();}
    
    
    
    

8.秦九昭算法

C/C++ code

    
    
    
    

     
     
     
     
     #include 
     
     
     
     <
     
     
     
     math.h
     
     
     
     >
     
     
     
     
     
     
     
     float
     
     
     
      qin(
     
     
     
     float
     
     
     
      a[],
     
     
     
     int
     
     
     
      n,
     
     
     
     float
     
     
     
      x){ 
     
     
     
     float
     
     
     
      r
     
     
     
     =
     
     
     
     0
     
     
     
     ; 
     
     
     
     int
     
     
     
      i; 
     
     
     
     for
     
     
     
     (i
     
     
     
     =
     
     
     
     n;i
     
     
     
     >=
     
     
     
     0
     
     
     
     ;i
     
     
     
     --
     
     
     
     ) r
     
     
     
     =
     
     
     
     r
     
     
     
     *
     
     
     
     x
     
     
     
     +
     
     
     
     a[i]; 
     
     
     
     return
     
     
     
      r;}main(){ 
     
     
     
     float
     
     
     
      a[
     
     
     
     50
     
     
     
     ],x,r
     
     
     
     =
     
     
     
     0
     
     
     
     ; 
     
     
     
     int
     
     
     
      n,i; 
     
     
     
     do
     
     
     
      { printf(
     
     
     
     "
     
     
     
     Input frequency:
     
     
     
     "
     
     
     
     ); scanf(
     
     
     
     "
     
     
     
     %d
     
     
     
     "
     
     
     
     ,
     
     
     
     &
     
     
     
     n); } 
     
     
     
     while
     
     
     
     (n
     
     
     
     <
     
     
     
     1
     
     
     
     ); printf(
     
     
     
     "
     
     
     
     Input value:
     
     
     
     "
     
     
     
     ); 
     
     
     
     for
     
     
     
     (i
     
     
     
     =
     
     
     
     0
     
     
     
     ;i
     
     
     
     <=
     
     
     
     n;i
     
     
     
     ++
     
     
     
     ) scanf(
     
     
     
     "
     
     
     
     %f
     
     
     
     "
     
     
     
     ,
     
     
     
     &
     
     
     
     a[i]); printf(
     
     
     
     "
     
     
     
     Input frequency:
     
     
     
     "
     
     
     
     ); scanf(
     
     
     
     "
     
     
     
     %f
     
     
     
     "
     
     
     
     ,
     
     
     
     &
     
     
     
     x); r
     
     
     
     =
     
     
     
     qin(a,n,x); printf(
     
     
     
     "
     
     
     
     Answer:%f
     
     
     
     "
     
     
     
     ,r); getch();}
    
    
    
    

9.幂法

C/C++ code

    
    
    
    

     
     
     
     
     #include
     
     
     
     <
     
     
     
     stdio.h
     
     
     
     >
     
     
     
     #include
     
     
     
     <
     
     
     
     math.h
     
     
     
     >
     
     
     
     
     
     
     
     #define
     
     
     
      N 100
     
     
     
     
     
     
     
     #define
     
     
     
      e 0.00001
     
     
     
     
     
     
     
     #define
     
     
     
      n 3
     
     
     
     
     
     
     
     float
     
     
     
      x[n]
     
     
     
     =
     
     
     
     {
     
     
     
     0
     
     
     
     ,
     
     
     
     0
     
     
     
     ,
     
     
     
     1
     
     
     
     };
     
     
     
     float
     
     
     
      a[n][n]
     
     
     
     =
     
     
     
     {{
     
     
     
     2
     
     
     
     ,
     
     
     
     3
     
     
     
     ,
     
     
     
     2
     
     
     
     },{
     
     
     
     10
     
     
     
     ,
     
     
     
     3
     
     
     
     ,
     
     
     
     4
     
     
     
     },{
     
     
     
     3
     
     
     
     ,
     
     
     
     6
     
     
     
     ,
     
     
     
     1
     
     
     
     }};
     
     
     
     float
     
     
     
      y[n];main(){ 
     
     
     
     int
     
     
     
      i,j,k; 
     
     
     
     float
     
     
     
      xm,oxm; oxm
     
     
     
     =
     
     
     
     0
     
     
     
     ; 
     
     
     
     for
     
     
     
     (k
     
     
     
     =
     
     
     
     0
     
     
     
     ;k
     
     
     
     <
     
     
     
     N;k
     
     
     
     ++
     
     
     
     ) { 
     
     
     
     for
     
     
     
     (j
     
     
     
     =
     
     
     
     0
     
     
     
     ;j
     
     
     
     <
     
     
     
     n;j
     
     
     
     ++
     
     
     
     ) { y[j]
     
     
     
     =
     
     
     
     0
     
     
     
     ; 
     
     
     
     for
     
     
     
     (i
     
     
     
     =
     
     
     
     0
     
     
     
     ;i
     
     
     
     <
     
     
     
     n;i
     
     
     
     ++
     
     
     
     ) y[j]
     
     
     
     +=
     
     
     
     a[j][i]
     
     
     
     *
     
     
     
     x[i]; } xm
     
     
     
     =
     
     
     
     0
     
     
     
     ; 
     
     
     
     for
     
     
     
     (j
     
     
     
     =
     
     
     
     0
     
     
     
     ;j
     
     
     
     <
     
     
     
     n;j
     
     
     
     ++
     
     
     
     ) 
     
     
     
     if
     
     
     
     (fabs(y[j])
     
     
     
     >
     
     
     
     xm) xm
     
     
     
     =
     
     
     
     fabs(y[j]); 
     
     
     
     for
     
     
     
     (j
     
     
     
     =
     
     
     
     0
     
     
     
     ;j
     
     
     
     <
     
     
     
     n;j
     
     
     
     ++
     
     
     
     ) y[j]
     
     
     
     /=
     
     
     
     xm; 
     
     
     
     for
     
     
     
     (j
     
     
     
     =
     
     
     
     0
     
     
     
     ;j
     
     
     
     <
     
     
     
     n;j
     
     
     
     ++
     
     
     
     ) x[j]
     
     
     
     =
     
     
     
     y[j]; 
     
     
     
     if
     
     
     
     (fabs(xm
     
     
     
     -
     
     
     
     oxm)
     
     
     
     <
     
     
     
     e) { printf(
     
     
     
     "
     
     
     
     max:%f/n/n
     
     
     
     "
     
     
     
     ,xm); printf(
     
     
     
     "
     
     
     
     v[i]:/n
     
     
     
     "
     
     
     
     ); 
     
     
     
     for
     
     
     
     (k
     
     
     
     =
     
     
     
     0
     
     
     
     ;k
     
     
     
     <
     
     
     
     n;k
     
     
     
     ++
     
     
     
     ) printf(
     
     
     
     "
     
     
     
     %f/n
     
     
     
     "
     
     
     
     ,y[k]); 
     
     
     
     break
     
     
     
     ; } oxm
     
     
     
     =
     
     
     
     xm; } getch();}
    
    
    
    

10.高斯塞德尔

C/C++ code

    
    
    
    

     
     
     
     
     #include
     
     
     
     <
     
     
     
     math.h
     
     
     
     >
     
     
     
     #include
     
     
     
     <
     
     
     
     stdio.h
     
     
     
     >
     
     
     
     
     
     
     
     #define
     
     
     
      N 20
     
     
     
     
     
     
     
     #define
     
     
     
      M 99
     
     
     
     
     
     
     
     float
     
     
     
      a[N][N];
     
     
     
     float
     
     
     
      b[N];
     
     
     
     int
     
     
     
      main(){ 
     
     
     
     int
     
     
     
      i,j,k,n; 
     
     
     
     float
     
     
     
      sum,no,d,s,x[N]; printf(
     
     
     
     "
     
     
     
     /nInput dim of n:
     
     
     
     "
     
     
     
     ); scanf(
     
     
     
     "
     
     
     
     %d
     
     
     
     "
     
     
     
     ,
     
     
     
     &
     
     
     
     n); 
     
     
     
     if
     
     
     
     (n
     
     
     
     >
     
     
     
     N) { printf(
     
     
     
     "
     
     
     
     Faild! Check if 0<n<N!/n 
     
     
     
     "
     
     
     
     ); getch(); 
     
     
     
     return
     
     
     
      
     
     
     
     1
     
     
     
     ; } 
     
     
     
     if
     
     
     
     (n
     
     
     
     <=
     
     
     
     0
     
     
     
     ) { printf(
     
     
     
     "
     
     
     
     Faild! Check if 0<n<N!/n 
     
     
     
     "
     
     
     
     );getch();
     
     
     
     return
     
     
     
      
     
     
     
     1
     
     
     
     ;} printf(
     
     
     
     "
     
     
     
     Input a[i,j],i,j=0…%d:/n
     
     
     
     "
     
     
     
     ,n
     
     
     
     -
     
     
     
     1
     
     
     
     ); 
     
     
     
     for
     
     
     
     (i
     
     
     
     =
     
     
     
     0
     
     
     
     ;i
     
     
     
     <
     
     
     
     n;i
     
     
     
     ++
     
     
     
     ) 
     
     
     
     for
     
     
     
     (j
     
     
     
     =
     
     
     
     0
     
     
     
     ;j
     
     
     
     <
     
     
     
     n;j
     
     
     
     ++
     
     
     
     ) scanf(
     
     
     
     "
     
     
     
     %f
     
     
     
     "
     
     
     
     ,
     
     
     
     &
     
     
     
     a[i][j]); printf(
     
     
     
     "
     
     
     
     Input b[i],i=0…%d:/n
     
     
     
     "
     
     
     
     ,n
     
     
     
     -
     
     
     
     1
     
     
     
     ); 
     
     
     
     for
     
     
     
     (i
     
     
     
     =
     
     
     
     0
     
     
     
     ;i
     
     
     
     <
     
     
     
     n;i
     
     
     
     ++
     
     
     
     ) scanf(
     
     
     
     "
     
     
     
     %f
     
     
     
     "
     
     
     
     ,
     
     
     
     &
     
     
     
     b[i]); 
     
     
     
     for
     
     
     
     (i
     
     
     
     =
     
     
     
     0
     
     
     
     ;i
     
     
     
     <
     
     
     
     n;i
     
     
     
     ++
     
     
     
     ) x[i]
     
     
     
     =
     
     
     
     0
     
     
     
     ; k
     
     
     
     =
     
     
     
     0
     
     
     
     ; printf(
     
     
     
     "
     
     
     
     /nk=%dx=
     
     
     
     "
     
     
     
     ,k); 
     
     
     
     for
     
     
     
     (i
     
     
     
     =
     
     
     
     0
     
     
     
     ;i
     
     
     
     <
     
     
     
     n;i
     
     
     
     ++
     
     
     
     ) printf(
     
     
     
     "
     
     
     
     %12.8f
     
     
     
     "
     
     
     
     ,x[i]); 
     
     
     
     do
     
     
     
      { k
     
     
     
     ++
     
     
     
     ; 
     
     
     
     if
     
     
     
     (k
     
     
     
     >
     
     
     
     M){printf(
     
     
     
     "
     
     
     
     /nError!/n”);getch();}
     
     
     
     
     
     
     
      
     
     
     
     break
     
     
     
     ; } no
     
     
     
     =
     
     
     
     0.0
     
     
     
     ; 
     
     
     
     for
     
     
     
     (i
     
     
     
     =
     
     
     
     0
     
     
     
     ;i
     
     
     
     <
     
     
     
     n;i
     
     
     
     ++
     
     
     
     ) { s
     
     
     
     =
     
     
     
     x[i]; sum
     
     
     
     =
     
     
     
     0.0
     
     
     
     ; 
     
     
     
     for
     
     
     
     (j
     
     
     
     =
     
     
     
     0
     
     
     
     ;j
     
     
     
     <
     
     
     
     n;j
     
     
     
     ++
     
     
     
     ) 
     
     
     
     if
     
     
     
      (j
     
     
     
     !=
     
     
     
     i) sum
     
     
     
     =
     
     
     
     sum
     
     
     
     +
     
     
     
     a[i][j]
     
     
     
     *
     
     
     
     x[j]; x[i]
     
     
     
     =
     
     
     
     (b[i]
     
     
     
     -
     
     
     
     sum)
     
     
     
     /
     
     
     
     a[i][i]; d
     
     
     
     =
     
     
     
     fabs(x[i]
     
     
     
     -
     
     
     
     s); 
     
     
     
     if
     
     
     
      (no
     
     
     
     <
     
     
     
     d) no
     
     
     
     =
     
     
     
     d; } printf(
     
     
     
     "
     
     
     
     /nk=%2dx=
     
     
     
     "
     
     
     
     ,k); 
     
     
     
     for
     
     
     
     (i
     
     
     
     =
     
     
     
     0
     
     
     
     ;i
     
     
     
     <
     
     
     
     n;i
     
     
     
     ++
     
     
     
     ) printf(
     
     
     
     "
     
     
     
     %f
     
     
     
     "
     
     
     
     ,x[i]);}
     
     
     
     while
     
     
     
      (no
     
     
     
     >=
     
     
     
     0.1e-6
     
     
     
     );
     
     
     
     if
     
     
     
     (no
     
     
     
     <
     
     
     
     0.1e-6
     
     
     
     ){ printf(
     
     
     
     "
     
     
     
     /n/n answer=/n
     
     
     
     "
     
     
     
     ); printf(
     
     
     
     "
     
     
     
     /nk=%d
     
     
     
     "
     
     
     
     ,k); 
     
     
     
     for
     
     
     
      (i
     
     
     
     =
     
     
     
     0
     
     
     
     ;i
     
     
     
     <
     
     
     
     n;i
     
     
     
     ++
     
     
     
     ) printf(
     
     
     
     "
     
     
     
     /n x[%d]=%12.8f
     
     
     
     "
     
     
     
     ,i,x[i]);}getch();} 
    
    
    
    

注:本文转自：http://topic.csdn.net/u/20080509/13/d4500a19-10aa-4fdf-b876-8e30f4daa331.html?seed=880334510