计算方法重要算法

1、#include#include#include#include#include#includeusing namespace std;void main()( int n;char c;void sub11(),sub22(),sub33(),sub44(),sub55(),sub66(),sub77();do(system(cls);coutnnn;coutn 1.数的合法输入并相加;coutn 2.线性插值之拉格朗日法”;coutn 3.二次插值(抛物线插值)之牛顿插值法”;coutn 4.n次插值之拉格朗日法;coutn 5.分段线性插值;coutn 6.最小二乘;coutn 7.高斯

2、消去;coutn 8. 退出”;coutn Please Input your choice!(1-8)n;scanf(%d,&n);switch (n)case 1: sub11( ); break;case 2: sub22( ); break;case 3: sub33( ); break;case 4: sub44( ); break;case 5: sub55( ); break;case 6: sub66( ); break;case 7: sub77( ); break;case 8: goto end;default : coutn 输入有误n;coutc;c=getchar(

3、); while(n!=8);end:;int is_number(const char *p)int flag0=0;int flag1=0;int flag3=0;while(*p)if (*p9)if (*p!=. & *p!=-) return -1;if (*p=.) flag0+;if (*p=-)if (flag30) return -1;else flag1+;if (flag0 1 | flag11 ) return -1;p+;flag3+;if (flag0=1)return 1; /floatelsereturn 0 ;/integervoid sub11()char

4、s110,s210;char *p1,*p2;cout s1;cout s2;p1=s1;p2=s2;if (is_number(p1)=0 & is_number(p2)=0) coutnatof(p1)+atof(p2); elsecouts1s2;void sub22()float x0,y0,x1,y1,xm,x,ym;float s;coutplease input two points:(x0,y0),(x1,y1) and (x0!=x1)”x0y0 x1y1;ym=y0*(x-x1)/(x0-x1)+y1*(x-x0)/(x1-x0);s=(y1-y0)/(x1-x0);cou

5、tplease input xm:xm;coutym=(y0*(xm-x1)/(x0-x1)+y1*(xm-x0)/(x1-x0)endl;cout方程是:y-”y1”=”s”(x-”x1”)”endl;void sub33()float x0,y0,x1,y1,x2,y2,xm,x,ym;float y01,y12,y012;coutplease input two points:(x0,y0),(x1,y1),(x2,y2)and (x0!=x1!=x2)”x0y0 x1y1x2y2;y01=(y0-y1)/(x0-x1);y12=(y1-y2)/(x1-x2);y012=(y01-y12

6、)/(x0-x2);ym=y0+(x-x0)*y01+(x-x0)*(x-x1)*y012;coutplease input xm:xm;coutym= y0+(xm-x0)*y01+(xm-x0)*(xm-x1)*y012endl;void sub44()int n,i,j;double xm,ym=0;double x200,y200, l200;cout n;cout endl;cout xm;cout endl;for(i=0;i=n;i+)cout x i xi;cout endl;for(i=0;i=n;i+)cout y i yi;cout endl;for(i=0;i=n;i+

7、)double b=1,c=1;for(j=0;j=n;j+)if(i!=j)b=b*(xm-xj);for(j=0;j=n;j+)if(i!=j)c=c*(xi-xj);li=b/c;for(i=0;i=n;i+)ym=ym+yi*li;cout ymendl;void sub55()int i,n;double xm,ym;cout xm;cout endl;cout n;cout endl;cout ”那么你现在一一输入这些坐标n”;double x200,y200;for(i=0;in;i+)cout x i xi;cout endl;for(i=0;in;i+)cout y i yi

8、;cout endl;for (i = 0; in; i+)if (xi = xm & xm=xi+1)ym=(yi*(xm-xi+1)/(xi-xi+1)+yi+1*(xm-xi)/(xi+1-xi);coutym=ymendl;double s=(yi+1-yi)/(xi+1-xi);cout该段对应的方程是:y-”yi+1”=”s”(x-”xi+1”)”endl;cout其中 x 的区间为【”xi”，”xi+1”】 ”显然 xm=xm 在这一区间内endl;break;void sub66()int n,i;float a,b,xsum=0,ysum=0,squaresum=0,xy=0

9、,x25,y25,D;coutn;cout请输入所有点坐标:n;for(i=0;ixiyi;xsum+=xi;ysum+=yi;squaresum+=xi*xi;xy+=xi*yi;D=n*squaresum-xsum*xsum;if(D=0)cout原方程无解或有多组解!n; elsea=(squaresum*ysum-xsum*xy)/D;b=(n*xy-xsum*ysum)/D;couty=a+bxn”;void sub77() cout测试例子：endl;couta11 = 4a12 = 1a13 = 1a14 = 1endl;couta21 = 1a22 = -1a23 = 2a24

10、 = -2endl;couta31 = 2a32 = 2coutvv”此方程组的解为：vvendl;a33 = -1a34 = 2endl;coutx1 = 1x=-1x3 = -2 endl;coutendl;int n;cout n;cout endl;cout ”请输入n行”n+1”列数据:endl;cout 注：其中每一行的最后一个值代表b1b2.bnendl;float *a ;a=new float*n+1;int i,j;for(i=1;i=n;i+)ai=new floatn+l;for(j=l;jv=n+l;j+)(cout a i j= cin aij;cout endl; 输入数据for(i=l ;iv=n;i+)(if(aii=0)(cout无解！ endl;goto end;float t;for(i=l ;iv=n;i+)(t=l/aii;for(int 1= 1 ;lv=n+1;1+) ail=t*ail;for(int m=i+1 ;mt*aij-amj; 转化过程coutendl转化为上三角矩阵：endl;for( i=l;iv=n;i+)(for(j=l;jv