10个重要的算法C语言实现源代码(留着以后看)

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

 

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-->#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(n*sizeof(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

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-->#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(n*sizeof(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];
 }
 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

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-->#include<stdio.h>
#include <math.h>
#define N 20
int main()
{ int n,i,j,k;
 int mi,tmp,mx;
 float a[N][N],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(a[i][j]);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]/a[n-1][n-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 A[NUMBER][NUMBER+1] ,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]=A[n][n+1]/A[n][n];
   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();}
 }
}

4.龙贝格求积公式，求解定积分

C/C++ code

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-->#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 T[MAX+1][2];
   T[0][1]=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

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-->5.牛顿迭代公式，求方程的近似解

C/C++ code

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-->#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 x*x*x+x*x-3*x-3; }
float f1(float x)
{ return 3.0*x*x+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();
}

6. 牛顿-科特斯求积公式，求定积分

C/C++ code

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-->#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

  


  

    
    
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2010-08-26 08:49
      
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