10个重要的算法C语言实现源代码

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(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

#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

#include<stdio.h>#include <math.h>#define N 20int 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 0x0dfloat 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

#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

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-5float 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

#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;elsenn=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 0.00001int 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 3float 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 99float 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();}