數(shù)值計(jì)算C語言常用小程序

1、1、 秦九韶算法 2、二分法 3、拉格朗日插值 4、埃特金算法 5、復(fù)化梯形法6、復(fù)化辛甫生算法 7、二階龍格庫(kù)塔方法 8、四階龍格庫(kù)塔方法9、改進(jìn)的歐拉方法10、迭代法11、埃特金加速方法：12、牛頓迭代法13、追趕法14、雅克比迭代15、蛋白質(zhì)設(shè)計(jì)：17高斯消去法: 1、 秦九韶算法P11.3 利用秦九韶算法求多項(xiàng)式 ，在x=3時(shí)的值。程序：#include<stdio.h>#include<math.h>void main()float a100,v,x;int n,i,k;scanf("%d%f",&n,&x);fo

2、r(i=0;i<=n;i+)scanf("%f",&ai);v=an;k=1;dov=x*v+an-k;k=k+1;while(k<=n);printf("v=%f",v); 運(yùn)行結(jié)果： 2、二分法P11.1用二分法求方程法x*x*x-x-1=0在1,2內(nèi)的近似根，要求誤差不超過#include<stdio.h>#include<math.h>float fun(float);void main()float a,b,c,x,y,y1;scanf("%f%f%f",&am

3、p;a,&b,&c);y1=fun(a);dox=(a+b)/2;y=fun(x);if(y*y1>0)a=x;elseb=x;while(b-a)>=c);printf("%f,%fn",x,y);float fun(float m)float n;n=m*m*m-m-1;return(n); 運(yùn)行結(jié)果：  3、拉格朗日插值程序：#include<stdio.h>main()float a,b,t,x100,y100;int n,i,j,k;scanf("%f%d",&a,&

4、amp;n);for(i=0;i<=n;i+)scanf("%f%f",&xi,&yi);k=0;b=0;for(k=0;k<=n;k+)t=1;for(j=0;j<=n;j+)if(j!=k)t=t*(a-xj)/(xk-xj);b=b+t*yk;printf("%fn",b);  4、埃特金算法程序：#include<stdio.h>#include<math.h>main()float a,b,c,x100,y100;int i,j,n,k;scanf("%d%

5、f",&n,&a);for(i=0;i<=n;i+)scanf("%f%f",&xi,&yi);for(k=1;k<=n;k+)for(i=k;i<=n;i+)yi=yk-1+(yi-yk-1)*(a-xk-1)/(xi-xk-1);printf("%fn",yn);5、復(fù)化梯形法P95.9 設(shè) ,用復(fù)化梯形法求積分 的近似值程序：#include<stdio.h>#include<math.h>double fun(double);void main()double a

6、,b,h,s,x,y;int n,k;scanf("%lf%lf%d",&a,&b,&n);h=(b-a)/n;s=0;x=a;y=fun(x);for(k=1;k<=n;k+)s=s+fun(x);x=x+h;s=s+fun(x);s=(h/2)*s;printf("s=%lfn",s);double fun(double m)double n;n=exp(-m)*sin(4*m)+1;return(n);運(yùn)行結(jié)果：  6、復(fù)化辛甫生算法P95.9 設(shè) ,用復(fù)化辛甫生法求積分 的近似值程序：#inclu

7、de<stdio.h>#include<math.h>double fun(double);void main()double a,b,h,s,x,y;int n,k;scanf("%lf%lf%d",&a,&b,&n);h=(b-a)/n;s=0;x=a;y=fun(x);for(k=1;k<=n;k+)s=s+fun(x);x=x+h/2;s=s+4*fun(x);x=x+h/2;s=s+fun(x);s=(h/6)*s;printf("s=%lfn",s);double fun(double m

8、)double n;n=exp(-m)*sin(4*m)+1;return(n);運(yùn)行結(jié)果：   7、二階龍格庫(kù)塔方法求解初值問題： 取h=0.2程序：#include<stdio.h>#include<math.h>float fun(float,float);void main()float h,x0,y0,x1,y1,k1,k2;int n,N;scanf("%f%f%f%d",&x0,&y0,&h,&N);n=1;for(n=1;n<=N;n+)x1=x0+h;k1

9、=fun(x0,y0);k2=fun(x0+h/2,y0+h/2*k1);y1=y0+h*k2;printf("%f,%fn",x1,y1);x0=x1;y0=y1;float fun(float a,float b)float m;m=b-2*a/b;return(m);運(yùn)行結(jié)果：   8、四階龍格庫(kù)塔方法求解初值問題： 取h=0.2程序：#include<stdio.h>#include<math.h>float fun(float,float);void main()float h,x0,y0,x1,y1

10、,k1,k2,k3,k4;int n,N;scanf("%f%f%f%d",&x0,&y0,&h,&N);n=1;for(n=1;n<=N;n+)x1=x0+h;k1=fun(x0,y0);k2=fun(x0+h/2,y0+h/2*k1);k3=fun(x0+h/2,y0+h/2*k2);k4=fun(x1,y0+h*k3);y1=y0+h/6*(k1+2*k2+2*k3+k4);printf("%f,%f",x1,y1);x0=x1;y0=y1;float fun(float a,float b)float m;m

11、=b-2*a/b;return(m);運(yùn)行結(jié)果：  9、改進(jìn)的歐拉方法求解初值問題： 程序：#include<stdio.h>#include<math.h>float fun(float,float);void main()float x0,y0,h,x1,y1,yp,yc;int n,N;scanf("%f%f%f%d",&x0,&y0,&h,&N);for(n=1;n<=N;n+)x1=x0+h;yp=y0+h*fun(x0,y0);yc=y0+h*fun(x1,yp);y1=

12、(yp+yc)/2;printf("%f,%f",x1,y1);x0=x1;y0=y1;float fun(float a,float b)float m;m=b-2*a/b;return(m);運(yùn)行結(jié)果：   10、迭代法P131 例2 用迭代法求方程 在 附近的一個(gè)根 ，要求精度為程序：#include<stdio.h>#include<math.h>float fun(float);void main()float x0,x1,c; int k,N; scanf("%f%f%d&quo

13、t;,&x0,&c,&N); for(k=1;k<=N;k+) x1=fun(x0);  printf("%fn",x1);  if(fabs(x1-x0)<c) break;  x0=x1; if(k-1=N)printf("Failure!n");float fun(float m)float n;n=exp(-m);return(n); 運(yùn)行結(jié)果：  11、埃特金加速方法：程序：#include<stdio.h>

14、#include<math.h>float fun(float);void main()float x0,x1,x2,c; int k,N; scanf("%f%f%d",&x0,&c,&N); for(k=1;k<=N;k+) x1=fun(x0);  x2=fun(x1);  x2=x2-(x2-x1)*(x2-x1)/(x2-2*x1+x0);  if(fabs(x2-x0)<c)  printf("%fn",x2);&#

15、160; break;  x0=x2; if(k-1=N)printf("Failure!n");float fun(float m)float n;n=exp(-m);return(n);運(yùn)行結(jié)果：  12、牛頓迭代法：例5、用牛頓法解方程牛頓公式為： ，取x=0.5程序：#include<stdio.h>#include<math.h>float fun(float);float ff(float);void main()float x0,x1,c; int k,N; scanf(&qu

16、ot;%f%f%d",&x0,&c,&N); for(k=1;k<=N;k+) if(ff(x0)!=0) x1=x0-fun(x0)/ff(x0); printf("%fn",x1);  if(fabs(x1-x0)<c) break;  x0=x1;  else        printf("*"); if(k=N)printf("Failure!

17、n");float fun(float m)float n;n=m-exp(-m);return(n);float ff(float m)float n;n=1+m;return(n);運(yùn)行結(jié)果：  13、追趕法： P198.3.用追趕法求解下列方程組：  程序：#include<stdio.h>#include<math.h>void main()float a100,b100,c100,d100,t;int i,n;scanf("%d",&n);for(i=2;i<=n;i

18、+)scanf("%f",&ai);for(i=1;i<=n;i+)scanf("%f",&bi);for(i=1;i<=n-1;i+)scanf("%f",&ci);for(i=1;i<=n;i+)scanf("%f",&di);c1=c1/b1;d1=d1/b1;for(i=2;i<n;i+)t=bi-ci-1*ai;ci=ci/t;di=(di-di-1*ai)/t;dn=(dn-dn-1*an)/(bn-cn-1*an);printf("%f

19、n",dn);for(i=n-1;i>=1;i-)di=di-ci*di+1;printf("%fn",di); 運(yùn)行結(jié)果：14、雅克比迭代程序：#include<stdio.h>#include<math.h>#define N 50                        

20、60;    #define M 4void main()double xM,yM,aMM,bM,dM,c,t; double ff(double ,int); int k,i,j,n;n=M-1;scanf("%lf",&c);                       

21、60;             for(i=0;i<=n;i+)scanf("%lf",&xi); scanf("%lf",&bi);   for(i=0;i<=n;i+)for(j=0;j<=n;j+)scanf("%lf",&a0i*M+j);for(k=1;k<=N;k+)for(i=1;i<=n;i+)fo

22、r(j=1,t=0;j<=n;j+)if(j=i)continue;elset=t+aij*xj;yi=(bi-t)/aii;for(i=1;i<=n;i+)di=fabs(xi-yi);t=ff(d,n+1);if(t<c) break;elsefor(i=1;i<=n;i+)xi=yi;if(k=N)printf("Failure!n");if(k<N)printf("k=%dn",k);for(i=1;i<=n;i+)printf("yi=%fn",yi);double ff(double a

23、,int n) double p;  int t;  p=a1; for(t=2;t<n;t+)if(p<at)p=at;return(p);運(yùn)行結(jié)果：   15、蛋白質(zhì)設(shè)計(jì)： 程序：#include <stdio.h>#include <stdlib.h>#include <math.h> void main()printf("橫坐標(biāo)     縱坐標(biāo)    豎坐標(biāo)n"

24、);  FILE *fp;char c100,x30007,y30007,z30007;float a3000,b3000,d3000,r3000;int i,k=0,j,n=0,m,p;float X3000,Y3000,Z3000,s1=0,s2=0,s3=0,av_x,av_y,av_z;fp=fopen("1a1c.pdb","r");for(i=0;i<=10000;i+)fgets(c,81,fp);if(c0='A'&&c1='T'&&c2='

25、O'&&c3='M')  for(j=0;j<6;j+)xkj=c32+j;                               printf("%c",xkj);printf("

26、60;   ");for(j=0;j<6;j+)                                ykj=c40+j;         &#

27、160;                     printf("%c",ykj);printf("    ");for(j=0;j<6;j+)              &

28、#160;                zkj=c48+j;printf("%c",zkj);printf("    ");Xk=atof(xk);s1+=Xk;Yk=atof(yk);s2+=Yk;Zk=atof(zk);s3+=Zk;printf("n");k+;if(c77='C') n+;av_x=s1/k;av_y=s2

29、/k;av_z=s3/k;printf("共有:%dn",k+1);printf("av_x=%f,av_y=%f,av_z=%fn",av_x,av_y,av_z);printf("C原子個(gè)數(shù)為:%dn",n);printf("平移原點(diǎn)后的坐標(biāo):n");for(m=0;m<k;m+)am=Xm-av_x;printf("%f     ",am);