牛頓迭代法解三元二次方程組(C++版)

1、牛頓迭代法解三元二次方程組（C+版） *方程組為：Y1=-X1+0.3X2*X3-X32+0.6=0Y2=-0.1X12-X2 +0.8X1*X3-X3+0.4=0Y3=0.3X1-0.5X22 +0.7X1*X2-X3+0.5=0 * #include#include#include #include#include #define N 3 / 非線性方程組中變量個數(shù)及方程個數(shù)const int N2=N*N; / jacobi矩陣的元素個數(shù)#define eps 0.00001 / 收斂精度#define Max 2000 / 最大迭代次數(shù)using namespace std;doubl

2、e main()double x0N,yN,x1N,es,esmax,jacobiNN,aij;int i,j,k,it=0,iter=0; ofstream fpout1(d:Program FilesMicrosoft Visual StudioMyProjectsxieFCZx0.txt,ios:out); if(!fpout1)coutfpout open fail!endl;return 0; /打開文件失敗則結(jié)束運行ifstream fpin(d:Program FilesMicrosoft Visual StudioMyProjectsxieFCZX0.txt,ios:in);i

3、f(!fpin)coutfpin open fail!endl;return 0; /打開文件失敗則結(jié)束運行 ofstream fpout2(d:Program FilesMicrosoft Visual StudioMyProjectsxieFCZanswer.txt,ios:out); if(!fpout2)coutfpout open fail!endl;return 0; /打開文件失敗則結(jié)束運行cout* Please input X0 *endl; / 技巧一：為保證jacobi矩陣盡快收斂，初值最好在 0 附近for(i=0;ix0i;cout輸入的初值為：endl;for(i=

4、0;iN;i+)coutx0it;coutendl;fpout1輸入的初值為：endl;do it+;for(i=0;iN;i+)fpout1x0it;fpout1endl;/jacobi have N2 element /計算jacobi矩陣的值jacobi00=0;jacobi01=0.3*x02;jacobi02=0.3*x01-2*x02;jacobi10=-0.2*x00+0.8*x02;jacobi11=0;jacobi12=0.8*x00-1;jacobi20=0.3+0.7*x01;jacobi21=-x01+0.7*x00;jacobi22=0;fpout2第 it 次jac

5、obi矩陣是：endl;for(i=0;iN;i+) for(j=0;jN;j+) fpout2.setf(cout.showpoint); /固定小數(shù)位數(shù)輸出fpout2setprecision(8)jacobiijt;fpout2endl;fpout2endl;aij=0;for(i=0;iN;i+) /計算jacobi矩陣元素的平方和aijfor(j=0;j1)cout1endl;k=rand()%3; x0k=(double)rand()/RAND_MAX;coutendl由計算機第 it 次隨機產(chǎn)生初值：endl;fpout1endl由計算機第 it 次隨機產(chǎn)生初值：endl;cou

6、t ;for(i=0;iN;i+) coutx0it; cout1); /保證aij1for(i=0;iN;i+)x1i=x0i;fpout2* 牛頓迭代如下 *endl;fpout2迭代次數(shù) x1tt x2tt x3endlendl;doiter=iter+1;fpout2 iter t;for(i=0;iN;i+)fpout2x1it;fpout2endl;/迭代格式y(tǒng)0=0.3*x11*x12-x12*x12+0.6; y1=-0.1*x10*x10+0.8*x10*x12-x12+0.4;y2=0.3*x10-0.5*x11*x11+0.7*x10*x11+0.5;/技巧二：由于要使最

7、終的aij小于0，可讓方程同除某個數(shù)使其系數(shù)小于1esmax=0.0;for(i=0;ifabs(esmax)esmax=es;if(fabs(esmax)eps)fpout2endlendl方程組的解為 : endl;coutendlendl方程組的解為 : endl;for(i=0;iN;i+)coutx1it;fpout2x1it;coutendlendl;fpout2endl;break;for(i=0;iN;i+)x1i=yi;while(iterMax);fpin.close();fpout1.close();fpout2.close();return 0;（讀入文件為 x0.txt，輸出文件