10个最重要的算法包括
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
/*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;}int 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();return 1;}/************************************************************* 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;}int 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();return 0;}
/*************************************************************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 0.00001float 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++ code5.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;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 0.00001int 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();}
，问候不一定要慎重其事，但一定要真诚感人
