包括拉格朗日，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 。都是经典的数学算法，希望能开托您的思路。转自kunli.info

1.拉格朗日插值多项式 ，用于离散数据的拟合

C/C++ code

include <stdio.h>

#include <conio.h>
#include <alloc.h>
float lagrange(float x,float y,float xx,int n) /拉格朗日插值算法/
{ int i,j;
float a,yy=0.0; /a作为临时变量，记录拉格朗日插值多项式*/
a=(float )malloc(nsizeof(float));
for(i=0;i<=n-1;i++)
{ a[i]=y[i];

 for(j=0;j<=n-1;j++)
 if(j!=i) a[i]*=(xx-x[j])/(x[i]-x[j]);
 yy+=a[i];

}
free(a);
return yy;
}
main()
{ int i,n;
float x[20],y[20],xx,yy;
printf("Input n:");
scanf("%d",&n);
if(n>=20) {printf("Error!The value of n must in (0,20)."); getch();return 1;}
if(n<=0) {printf("Error! The value of n must in (0,20)."); getch(); return 1;}
for(i=0;i<=n-1;i++)
{ printf("x[%d]:",i);

scanf("%f",&x[i]);

}
printf("\n");
for(i=0;i<=n-1;i++)
{ printf("y[%d]:",i);scanf("%f",&y[i]);}
printf("\n");
printf("Input xx:");
scanf("%f",&xx);
yy=lagrange(x,y,xx,n);
printf("x=%f,y=%f\n",xx,yy);
getch();
}

2.牛顿插值多项式，用于离散数据的拟合

C/C++ code

include <stdio.h>

include <conio.h>

include <alloc.h>

void difference(float x,float y,int n)
{ float *f;
int k,i;
f=(float )malloc(nsizeof(float));
for(k=1;k<=n;k++)
{ f[0]=y[k];

for(i=0;i<k;i++)
  f[i+1]=(f[i]-y[i])/(x[k]-x[i]);
y[k]=f[k];

}//代码效果参考：http://www.zidongmutanji.com/bxxx/414936.html

return;
}
main()
{ int i,n;
float x[20],y[20],xx,yy;
printf("Input n:");
scanf("%d",&n);
if(n>=20) {printf("Error! The value of n must in (0,20)."); getch(); return 1;}
if(n<=0) {printf("Error! The value of n must in (0,20).");getch(); return 1;}
for(i=0;i<=n-1;i++)
{ printf("x[%d]:",i);

scanf("%f",&x[i]);

}
printf("\n");
for(i=0;i<=n-1;i++)
{ printf("y[%d]:",i);scanf("%f",&y[i]);}
printf("\n");
difference(x,(float *)y,n);
printf("Input xx:");
scanf("%f",&xx);
yy=y[20];
for(i=n-1;i>=0;i--) yy=yy*(xx-x[i])+y[i];
printf("NewtonInter(%f)=%f",xx,yy);
getch();
}

3.高斯列主元消去法,求解其次线性方程组

C/C++ code

include<stdio.h>

include <math.h>

define N 20

int main()
{ int n,i,j,k;
int mi,tmp,mx;
float aN,b[N],x[N];
printf("\nInput n:");
scanf("%d",&n);
if(n>N)
{ printf("The input n should in(0,N)!\n");

getch();
return 1;

}
if(n<=0)
{ printf("The input n should in(0,N)!\n");

getch();
return 1;

}
printf("Now 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("Now input b(i),i,j=0...%d:\n",n-1);
for(i=0;i<n;i++)
scanf("%f",&b[i]);
for(i=0;i<n-2;i++)
{ for(j=i+1,mi=i,mx=fabs(ai);j<n-1;j++)

if(fabs(a[j][i])>mx)
{ mi=j;
  mx=fabs(a[j][i]);
}
if(i<mi)
{ tmp=b[i];b[i]=b[mi];b[mi]=tmp;
  for(j=i;j<n;j++)
  { tmp=a[i][j];
    a[i][j]=a[mi][j];
    a[mi][j]=tmp;
  }
}
for(j=i+1;j<n;j++)
{ tmp=-a[j][i]/a[i][i];
  b[j]+=b[i]*tmp;
  for(k=i;k<n;k++)
  a[j][k]+=a[i][k]*tmp;
}

}
x[n-1]=b[n-1]/an-1;
for(i=n-2;i>=0;i--)
{ x[i]=b[i];

for(j=i+1;j<n;j++)
x[i]-=a[i][j]*x[j];
x[i]/=a[i][i];

}
for(i=0;i<n;i++)
printf("Answer:\n x[%d]=%f\n",i,x[i]);
getch();
return 0;
}

include<math.h>

include<stdio.h>

define NUMBER 20

define Esc 0x1b

define Enter 0x0d

float ANUMBER ,ark;
int flag,n;
exchange(int r,int k);
float max(int k);
message();

main()
{
float x[NUMBER];
int r,k,i,j;
char celect;
clrscr();

printf("\n\nUse Gauss.");
printf("\n\n1.Jie please press Enter.");
printf("\n\n2.Exit press Esc.");
celect=getch();
if(celect==Esc)

 exit(0);

printf("\n\n input n=");
scanf("%d",&n);

 printf(" \n\nInput matrix A and B:");

for(i=1;i<=n;i++)
{

printf("\n\nInput a%d1--a%d%d and b%d:",i,i,n,i);

for(j=1;j<=n+1;j++)        scanf("%f",&A[i][j]);

}
for(k=1;k<=n-1;k++)
{
ark=max(k);

if(ark==0)
{
  printf("\n\nIt's wrong!");message();
}
else if(flag!=k)
 exchange(flag,k);
 for(i=k+1;i<=n;i++)
 for(j=k+1;j<=n+1;j++)
 A[i][j]=A[i][j]-A[k][j]*A[i][k]/A[k][k];

}
x[n]=An/An;
for( k=n-1;k>=1;k--)
{

 float me=0;
 for(j=k+1;j<=n;j++)
 {
   me=me+A[k][j]*x[j];
 }
   x[k]=(A[k][n+1]-me)/A[k][k];

}
for(i=1;i<=n;i++)
{

 printf(" \n\nx%d=%f",i,x[i]);

}
message();
}

exchange(int r,int k)
{
int i;
for(i=1;i<=n+1;i++)

A[0][i]=A[r][i];

for(i=1;i<=n+1;i++)

A[r][i]=A[k][i];

for(i=1;i<=n+1;i++)

A[k][i]=A[0][i];

}

float max(int k)
{
int i;
float temp=0;
for(i=k;i<=n;i++)

if(fabs(A[i][k])>temp)
{
  temp=fabs(A[i][k]);
  flag=i;
}

return temp;
}

message()
{
printf("\n\n Go on Enter ,Exit press Esc!");
switch(getch())
{
case Enter: main();
case Esc: exit(0);
default:{printf("\n\nInput error!");message();}
}
}
//代码效果参考：http://www.zidongmutanji.com/zsjx/332439.html

4.龙贝格求积公式，求解定积分

C/C++ code

include<stdio.h>

include<math.h>

define f(x) (sin(x)/x)

define N 20

define MAX 20

define a 2

define b 4

define e 0.00001

float LBG(float p,float q,int n)
{ int i;
float sum=0,h=(q-p)/n;
for (i=1;i<n;i++)
sum+=f(p+i*h);
sum+=(f(p)+f(q))/2;
return(h*sum);
}
void main()
{ int i;
int n=N,m=0;
float TMAX+1;
T0=LBG(a,b,n);
n*=2;
for(m=1;m<MAX;m++)
{ for(i=0;i<m;i++)

  T[i][0]=T[i][1];
 T[0][1]=LBG(a,b,n);
 n*=2;
 for(i=1;i<=m;i++)
 T[i][1]=T[i-1][1]+(T[i-1][1]-T[i-1][0])/(pow(2,2*m)-1);
 if((T[m-1][1]<T[m][1]+e)&&(T[m-1][1]>T[m][1]-e))
 { printf("Answer=%f\n",T[m][1]); getch();
  return ;
 }

}
}

C/C++ code
5.牛顿迭代公式，求方程的近似解

C/C++ code

include<stdio.h>

include<math.h>

include<conio.h>

define N 100

define PS 1e-5

define TA 1e-5

float Newton(float (f)(float),float(f1)(float),float x0 )
{ float x1,d=0;
int k=0;
do
{ x1= x0-f(x0)/f1(x0);

if((k++>N)||(fabs(f1(x1))<PS))
{ printf("\nFailed!");
  getch();
  exit();
}
d=(fabs(x1)<1?x1-x0:(x1-x0)/x1);
x0=x1;
printf("x(%d)=%f\n",k,x0);

}
while((fabs(d))>PS&&fabs(f(x1))>TA) ;
return x1;
}
float f(float x)
{ return xxx+xx-3x-3; }
float f1(float x)
{ return 3.0xx+2*x-3; }
void main()
{ float f(float);
float f1(float);
float x0,y0;
printf("Input x0: ");
scanf("%f",&x0);
printf("x(0)=%f\n",x0);
y0=Newton(f,f1,x0);
printf("\nThe root is x=%f\n",y0);
getch();
}

1. 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 xi:\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 aN,bN,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",&ai);
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++)
{ bi=-ai/ai; g[i]=c[i]/ai; }
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
//代码效果参考：http://www.zidongmutanji.com/zsjx/536177.html

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 an={{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 aN;
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",&ai);
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();
}