10个重要的算法C语言实现源代码

CocoaChina 2010-08-09 17:45:31 7369

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 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
#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-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
#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();
}