曲线拟合最小二乘法C++程序

1、课题八曲线拟合的最小二乘法实验目标:在某冶炼过程中，通过实验检测得到含碳量与时间关系的数据如下，试求含碳量y与时间t内在关系的拟合曲线。t05104550Y01.272.16:2.863.443.874.154.374.51P 4.584.02最小二乘法：#inelude <iostream>#inelude <math.h>using namespaeestd;int Array( double *Arr, int n)double *p;int i;p=(double *)malloc(n* sizeof (double *);if (!p) return 0;fo

2、r (i=0;i<n;i+)pi=( double *)malloc(n* sizeof (double); if (!pi) return 0;*Arr=p;return 1;void main()int n,i,j,k;double *A,*B,*S;S=new double 3;B=new double 3;coutvv"共2有？DN个?节 _2 点 i? e o" cin»n;eout<< "请？输 o?入？节 T点值 | i ( e ?§ X) e? e o<<endl;double *X;X = new

3、 double n;for (i=0;i<n;i+)ci n»Xi;eout<<"请？输o?入？节 T点i函？盘数oy值| i(e ? § Y) e? e o<<endl;double *Y;Y = new double n;for (i=0;i<n;i+)ei n»Yi;if (!Array(&A,3)eout<<"内 T 存？?分O?配?失o?败？'T ! e ?;elsefor (i=0;i<3;i+)for (j=0;j<3;j+)Aij=0;for (i=0;

4、i<n;i+)A00+=1; / eout<<A00;A01+=Xi;A02+=Xi*Xi;A10=A01;A11=A02;A12+=Xi*Xi*Xi;A20=A02;A21=A12;A22+=Xi*Xi*Xi*Xi;for (i=0;i<3;i+)Bi=0;for (i=0;i<n;i+)B0+=Yi;B1+=Xi*Yi;B2+=Xi*Xi*Yi;for ( k = 0; k < 2 ;k+)for (i = k+1; i<3; i+)double T = Aik/Akk;Bi = Bi - T * Bk;for ( j = k+1 ; j <

5、 3 ; j+ )Aij = Aij - T * Akj;S2 = B2/A22;for (i = 1; i >=0 ; i-)double Temp = 0;for ( int j = i+1; j<2 ;j+)Temp = Temp + Aij * Sj;Si = (Bi - Temp) /Aii;cout«"拟a合？曲"2 线？为a： e o<<endl;cout<< "y=" <<S0<< " + "<<S1<< " x +

6、"<<S2<<" xA2 " <<e ndl«e ndl<<"误"？差？为a： e o<<endl;for (i=0;i<n;i+)cout<< "Y" <<i<< "-Y(t" vvivv ")=" <<S0+Si*Xi +S2*Xi*Xi-Yi«e ndl;II®CUser5Appl eDe s IctopVS 育分听2 012.5 r28

7、zu ixia oDe b ugzu ixiao. exe11沁)：iR 5 10 15 20 25 30 35 £0 45 50惰输入节点函数值(Vi)=p 1.27 2-16 2.96 3-44 3-87 4-15 4.3? 4-51 4.SB 4.B2撕合曲线3:k/=l. 19727+ 0.08021S2 x + -0.00287459 xA2V0-V<t0>=1-19727Vl-¥<tl>=0,256499V2-¥<t2>=-l.27893V3-¥<t3)=-2.30951V4-V<t4>=-5-31397R+304V5-y<t