实验一：拉格朗日插值多项式命名（源程序.cpp及工作区.dsw）：lagrange问题：4//Lagrange.cpp#include <stdio.h>#include <conio.h>#define N 4int checkvalid(double x[], int n);void printLag (double x[], double y[], double varx, int n);double Lagrange(double x[], double y[], double varx, int n);void main (){double x[N+1] = {0.4, 0.55, 0.8, 0.9, 1};double y[N+1] = {0.41075, 0.57815, 0.88811, 1.02652, 1.17520};double varx = 0.5;if (checkvalid(x, N) == 1){printf("\n\n插值结果: P(%f)=%f\n", varx, Lagrange(x, y, varx, N));}else{printf("结点必须互异");}getch();}int checkvalid (double x[], int n){int i,j;for (i = 0; i < n; i++){for (j = i + 1; j < n+1; j++){if (x[i] == x[j])//若出现两个相同的结点，返回-1{return -1;}}}return 1;}double Lagrange (double x[], double y[], double varx, int n) {double fenmu;double fenzi;double result = 0;int i,j;printf("Ln(x) =\n");for (i = 0; i < n+1; i++){fenmu = 1;for (j = 0; j < n+1; j++){if (i != j){fenmu = fenmu * (x[i] - x[j]);}}printf("\t%f", y[i] / fenmu);fenzi = 1;for (j = 0; j < n+1; j++){if (i != j){printf("*(x-%f)", x[j]);fenzi = fenzi * (varx - x[j]);}}if (i != n){printf("+\n");}result += y[i] / fenmu * fenzi;}return result;}运行及结果显示：实验二：牛顿插值多项式命名（源程序.cpp及工作区.dsw）：newton_cz问题：4//newton_cz.cpp#include<stdio.h>#include<iostream.h>#include<conio.h>#define N 4int checkvaild(double x[],int n){int i,j;for(i=0;i<n+1;i++){for(j=i+1;j<n+1;j++)if(x[i]==x[j])return -1;}return 1;}void chashang(double x[],double y[],double f[][N+1]){int i,j,t=0;for(i=0;i<N+1;i++){f[i][0]=y[i];//f[0][0]=y[0],f[1][0]=y[1];f[2][0]=y[2];f[3][0]=y[3];f[4][0]=y[4] }for(j=1;j<N+1;j++)// 阶数j{for(i=0;i<N-j+1;i++) //差商个数if[i][j]=(f[i+1][j-1]-f[i][j-1])/(x[t+i+1]-x[i]);//一阶为f[0][1]~f[3][1];二阶为f[0][2]~f[2][2]//三阶为f[0][3]~f[1][3];四阶为f[0][4]t++;}}double compvalue(double t[][N+1],double x[],double varx){int i,j,r=0;double sum=t[0][0],m[N+1]={sum,1,1,1,1};printf("i\tXi\t F(Xi)\t 1阶\t 2阶\t\t3阶\t 4阶\n");printf("--------------------------------------------------------------------------------");for(i=0;i<N+1;i++){r=i;printf("x%d\t%f ",i,x[i]);for(j=0;j<=i;j++){printf("%f ",t[r][j]);r--;}printf("\n");}printf("--------------------------------------------------------------------------------");/**/ printf("N(x) =\n");printf(" %f\n",t[0][0]);for(i=1;i<N+1;i++){for(j=0;j<i;j++)m[i]=m[i]*(varx-x[j]);m[i]=t[0][i]*m[i];sum=sum+m[i];}for(i=1;i<N+1;i++){ printf(" +%f*",t[0][i]);for(j=0;j<i;j++)printf("(x-%f)",x[j]);printf("\n");}return sum;}void main(){double varx,f[N+1][N+1];double x[N+1]={0.4,0.55,0.8,0.9,1};double y[N+1]={0.41075,0.57815,0.88811,1.02652,1.17520};checkvaild(x,N);chashang(x,y,f);varx=0.5;if(checkvaild(x,N)==1){chashang(x,y,f);printf("\n\n牛顿插值结果: P(%f)=%f\n",varx,compvalue(f,x,varx));}elseprintf("输入的插值节点的x值必须互异!");getch();} 运行及结果显示：实验三：自适应梯形公式命名（源程序.cpp 及工作区.dsw ）：autotrap ][n T问题：计算⎰+=10214dx x π的近似值，使得误差（EPS ）不超过610- //autotrap.cpp #include<stdio.h> #include<conio.h> #include<math.h> #define EPS 1e-6 double f(double x) { return 4/(1+x*x); } double AutoTrap(double(*f)(double),double a,double b,double eps) { double t2,t1,sum=0.0; int i,k; t1=(b-a)*(f(a)+f(b))/2; printf("T(%4d)=%f\n",1,t1); for(k=1;k<11;k++) { for(i=1;i<=pow(2,k-1);i++) sum+=f(a+(2*i-1)*(b-a)/pow(2,k)); t2=t1/2+(b-a)*sum/pow(2,k); printf("T(%4d)=%f\n",(int)pow(2,k),t2); if(fabs(t2-t1)<EPS) break; else t1=t2; sum=0.0; } return sum; } void main(){ double s;s=AutoTrap(f,0.0,1.0,EPS);getch();} 运行及结果显示：实验四：龙贝格算法命名（源程序.cpp 及工作区.dsw ）：romberg问题：求⎰+=10214dx x π的近似值，要求误差不超过610-//romberg.cpp #include <stdio.h> #include <conio.h> #include <math.h> #define N 20 #define EPS 1e-6 double f(double x){return 4/(1+x*x);} double Romberg(double a,double b,double (*f)(double),double eps) { double T[N][N],p,h;int k=1,i,j,m=0;T[0][0]=(b-a)/2*(f(a)+f(b));do{p=0;h=(b-a)/pow(2,k-1);for(i=1;i<=pow(2,k-1);i++)p=p+f(a+(2*i-1)*h/2);T[0][k]=T[0][k-1]/2+p*h/2;for(i=1;i<=k;i++){j=k-i;T[i][j]=(pow(4,i)*T[i-1][j+1]-T[i-1][j])/(pow(4,i)-1); }k++; p=fabs(T[k-1][0]-T[k-2][0]);}while(p>=EPS);k--; while(m<=k){for(i=0;i<=m;i++) printf("T(%d,%d)=%f ",i,m-i,T[i][m-i]);m++;printf("\n");}return T[k][0]; } void main() {printf("\n\n 积分结果 = %f",Romberg(0,1,f,EPS)); getch(); } 运行及结果显示：实验五：牛顿切线法问题：求方程01)(3=--=x x x f 在5.1=x ，6.0=x 附近的根（精度31021-⨯=） //newton_qxf.cpp#include <math.h>#include<conio.h>#include <stdio.h>#define MaxK 1000#define EPS 0.5e-3double f(double x){return x*x*x-x-1;}double f1(double x){return 3*x*x-1;}int newton(double (*f)(double), double (*f1)(double), double &x0, double eps) {double xk, xk1;int count = 0;printf("k\txk\n");printf("-----------------------\n");xk = x0;printf("%d\t%f\n", count, xk);do{if((*f1)(xk)==0)return 2;xk1 = xk - (*f)(xk) / (*f1)(xk);if (fabs(xk - xk1) < eps){count++;xk = xk1;printf("%d\t%f\n", count, xk);x0 = xk1;return 1;}count++;xk = xk1;printf("%d\t%f\n", count, xk);}while(count < MaxK);return 0;}void main(){for(int i=0;i<2;i++){double x=0.6;if(i==1)x=1.5;printf("x0初值为%f:\n",x);if (newton(f, f1, x, EPS) == 1){printf("-----------------------\n");printf("the root is x=%f\n\n\n", x);}else{printf("the method is fail!");}}getch();}实验六：牛顿下山法命名（源程序.cpp 及工作区.dsw ）：newton_downhill 问题：求方程01)(3=--=x x x f 在5.1=x ，6.0=x 附近的根（精度31021-⨯=） //newton_downhill.cpp #include <stdio.h> #include <conio.h> #include <math.h> #include <stdlib.h> #define Et 1e-3//下山因子下界 #define E1 1e-3//根的误差限 #define E2 1e-3//残量精度 double f(double x) { return x * x * x - x - 1; } double f1(double x) { return 3 * x * x - 1; } void errormess(int b){char *mess;switch(b){case -1:mess = "f(x)的导数为0!";break;case -2:mess = "下山因子已越界，下山处理失败";break;default:mess = "其他类型错误!";}printf("the method has faild! because %s", mess);}int Newton(double (*f)(double), double (*f1)(double), double &x0) {int k = 0;double t;double xk, xk1;double fxk, fxk_temp;printf("k t xk f(xk)\n");printf("----------------------------------------------------------\n");printf("%-20d", k);xk = x0;printf("%-15f", x0);fxk = (*f)(xk);printf("%-20f", fxk);printf("\n");for (k = 1; ; k++){t = 1;while(1){printf("%-10d", k);printf("%-10f", t);if((*f1)(xk) != 0){xk1 = xk - t * (*f)(xk) / (*f1)(xk);}else{return -1;}printf("%-15f", xk1);fxk_temp = (*f)(xk1);printf("%-20f", fxk_temp);if(fabs(fxk_temp) >fabs(fxk)){t = t / 2;printf("\n");if (t < Et){return -2;}}else{printf("下山成功\n");break;}}if (fabs(xk-xk1)<E1){x0 = xk1;return 1; } xk = xk1; } }void main() {int b;double x0; x0 = 0.6;b = Newton(f, f1, x0); if (b == 1) printf("\nthe root x=%f\n", x0); else errormess(b); getch(); }运行及结果显示：实验七：埃特金加速算法命名（源程序.cpp 及工作区.dsw ）：aitken问题：求方程01)(3=--=x x x f 在5.1=x ，6.0=x 附近的根（精度31021-⨯=） //aitken.cpp#include <math.h> #include <stdio.h> #include <conio.h> #define MaxK 100 #define EPS 0.5e-3double g(double x) {return x * x * x - 1; }int aitken (double (*g)(double), double &x, double eps) {int k;double xk = x, yk, zk, xk1;printf("k xk yk zk xk+1\n");printf("-------------------------------------------------------------------\n"); for (k = 0;k<MaxK; k++) { yk = (*g)(xk); zk = (*g)(yk);xk1 = xk - (yk - xk) * (yk - xk) / (zk - 2 * yk + xk); printf("%-10d%-15f%-15f%-15f%-15f\n", k, xk, yk, zk, xk1); if (fabs(yk-xk)<=eps) { x = xk1; return k + 1; } xk = xk1; }return -1; }void main () {double x = 1.5; int k;k = aitken(g, x, EPS); if (k == -1) printf("迭代次数越界!\n"); else printf("\n 经k=%d 次迭代，所得方程根为:x=%f\n", k, x); getch(); }运行及结果显示：实验八：正割法问题：求方程01)(3=--=x x x f 在5.1=x 附近的根（精度0.5e-8） //ZhengGe.cpp #include <math.h> #include <stdio.h> #include <conio.h>#define MaxK 1000 #define EPS 0.5e-8double f(double x) {return x*x*x-x-1;}int ZhengGe(double (*f)(double), double x0, double &x1, double eps) {printf("k xk f(xk)\n");printf("---------------------------------------------\n");int k;double xk0, xk, xk1;xk0 = x0;printf("%-10d%-15f%-15f\n", 0, x0, (*f)(x0));xk = x1;for (k=1;k<MaxK;k++){if((*f)(xk)-(*f)(xk0)==0)return -1;xk1 = xk - (*f)(xk) / ( (*f)(xk) - (*f)(xk0) ) * (xk - xk0);printf("%-10d%-15f%-15f\n", k, xk, (*f)(xk));if (fabs(xk1 - xk)<=eps){printf("%-10d%-15f%-15f\n\n", k + 1, xk1, (*f)(xk1));printf("---------------------------------------------\n\n");x1 = xk1;return 1;}xk0 = xk;xk = xk1;}return -2;}void main (){double x0 = 1, x1 = 2;if (ZhengGe(f, x0, x1, EPS) == 1){printf("the root is x = %f\n", x1);}else{printf("the method is fail!");}getch();}实验九：高斯列选主元算法命名（源程序.cpp 及工作区.dsw ）：colpivot问题：求解方程组并计算系数矩阵行列式值 ⎪⎩⎪⎨⎧=-+=+-=-+2240532321321321x x x x x x x x x//colpivot.cpp#include <math.h> #include <stdio.h> #include <conio.h> #define N 3static double aa[N][N]={{1,2,-1},{1,-1,5},{4,1,-2}}; static double bb[N+1]={3,0,2};void main() {int i,j;double a[N+1][N+1],b[N+1],x[N+1],det;double gaussl(double a[][N+1],double b[],double x[]); for(i=1;i<=N;i++) { for(j=1;j<=N;j++) a[i][j]=aa[i-1][j-1]; b[i]=bb[i-1]; }det=gaussl(a,b,x); if(det!=0) { printf("\n 方程组的解为:"); for(i=1;i<=N;i++) printf(" x[%d]=%f",i,x[i]); printf("\n\n 系数矩阵的行列式值＝%f",det); }else printf("\n\n 系数矩阵奇异阵，高斯方法失败 !:"); getch(); }double gaussl(double a[][N+1],double b[],double x[])//a传入增广矩阵(有效元素为a[1,1]...a[n,n+1]),x负责传入迭代初值并返回解向量//(有效值为x[1]...x[n]);返回值：系数矩阵行列式值detA{double det=1.0,F,m,temp;int i,j,k,r;void disp(double a[][N+1],double b[]);printf("消元前增广矩阵:\n");disp(a,b);for(k=1;k<N;k++){temp=a[k][k];r=k;for(i=k+1;i<=N;i++){if(fabs(temp)<fabs(a[i][k])){temp=a[i][k];r=i;}//按列选主元，即确定ik}if(a[r][k]==0)return 0;//如果aik,k=0,则A为奇异矩阵，停止计算if(r!=k){for(j=k;j<=N;j++){a[k][j]+=a[r][j];a[r][j]=a[k][j]-a[r][j];a[k][j]-=a[r][j];}b[k]+=b[r];b[r]=b[k]-b[r];b[k]-=b[r];det=-det;//如果ik!=k,则交换[A,b]第ik行与第k行元素printf("交换第%d, %d行:\n",k,r);disp(a,b);}for(i=k+1;i<=N;i++){m=a[i][k]/a[k][k];for(j=1;j<=k;j++)a[i][j]=0.0;for(j=k+1;j<=N;j++)a[i][j]-=m*a[k][j];b[i]-=m*b[k];}printf("第%d次消元:\n",k);disp(a,b);det*=a[k][k];} //大FOR循环结束x[N]=b[N]/a[N][N];for(i=N-1;i>0;i--){F=0.0;for(j=i+1;j<=N;j++)F+=a[i][j]*x[j];x[i]=(b[i]-F)/a[i][i];}det*=a[N][N];return det;}void disp(double a[][N+1],double b[])//显示选主元及消元运算中间增广矩阵结果 {int i,j;for(i=1;i<=N;i++) {for(j=1;j<=N;j++)printf("%10f\t",a[i][j]); printf("%10f\n",b[i]); } }运行及结果显示：实验十：高斯全主元消去算法 命名（源程序.cpp 及工作区.dsw ）：fullpivot问题：利用全主元消去法求解方程组⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡----9555.04525.015.0201.0321x x x //fullpivot.cpp#include <math.h> #include <stdio.h> #include <conio.h> #define N 3 double x[N+1];static double aa[N][N]={{0.01,2,-0.5},{-1,-0.5,2},{5,-4,0.5}}; static double bb[N]={-5,5,9};void main(){int i,j;double a[N+1][N+1],b[N+1],x[N+1],det;int t[N+1];//引入列交换保存x向量的各分量的位置顺序double gaussl(double a[][N+1],double b[],double x[],int t[]);for(i=1;i<=N;i++){for(j=1;j<=N;j++) a[i][j]=aa[i-1][j-1];b[i]=bb[i-1];}for(i=1;i<=N;i++)t[i]=i;det=gaussl(a,b,x,t);if(det!=0){ printf("\n方程组的解为:");for(i=N;i>=1;i--){printf(" x[%d]=%f",t[i],x[i]);if(i>1)printf(" -->");}printf("\n\n系数矩阵的行列式值＝%f",det);}else printf("\n\n系数矩阵奇异阵，高斯方法失败!:");getch();}double gaussl(double a[][N+1],double b[],double x[],int t[])//a传入增广矩阵(有效元素为a[1,1]...a[n,n+1]),x负责传入迭代初值并返回解向量//(有效值为x[1]...x[n]);返回值：系数矩阵行列式值detA{double det=1.0,F,m,temp;int i,j,k,r,s;void disp(double a[][N+1],double b[],int x[]);// printf("消元前增广矩阵:\n");//disp(a,b,t);for(k=1;k<N;k++){temp=a[k][k];r=k;s=k;for(i=k;i<=N;i++)for(j=k;j<=N;j++)if(fabs(temp)<fabs(a[i][j])){temp=a[i][j];r=i;s=j;}//选主元，选取ik,jkif(a[r][s]==0)return 0;if(r!=k){for(j=k;j<=N;j++){a[k][j]+=a[r][j];a[r][j]=a[k][j]-a[r][j];a[k][j]-=a[r][j];}b[k]+=b[r];b[r]=b[k]-b[r];b[k]-=b[r];det=-det;printf("交换第%d, %d行:\n",k,r);disp(a,b,t);}if(s!=k){for(i=0;i<=N;i++){a[i][k]+=a[i][s];a[i][s]=a[i][k]-a[i][s];a[i][k]-=a[i][s];}t[k]+=t[s];t[s]=t[k]=t[s];t[k]-=t[s];det=-det;printf("交换第%d, %d列:\n",k,s);disp(a,b,t);}for(i=k+1;i<=N;i++){m=a[i][k]/a[k][k];for(j=1;j<=k;j++)a[i][j]=0.0;for(j=k+1;j<=N;j++)a[i][j]-=m*a[k][j];b[i]-=m*b[k];} //消元计算printf("第%d次消元:\n",k);disp(a,b,t);det*=a[k][k];} //大FOR循环结束x[N]=b[N]/a[N][N];for(i=N-1;i>0;i--){F=0.0;for(j=i+1;j<=N;j++)F+=a[i][j]*x[j];x[i]=(b[i]-F)/a[i][i];}//回代计算det*=a[N][N];return det;}void disp(double a[][N+1],double b[],int x[])//显示选主元及消元运算中间增广矩阵结果{int i,j;for(i=1;i<=N;i++){for(j=1;j<=N;j++)printf("%f\t",a[i][j]);printf("| x%d = %f\n",x[i],b[i]);}}运行及结果显示：实验十一：LU 分解算法命名（源程序.cpp 及工作区.dsw ）：LU问题：利用LU 法求解方程组 ⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡=⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡472111312613321x x x //LU.cpp#include<math.h> #include<stdio.h> #include<conio.h> #define N 3static aa[N][N]={{3,1,6},{2,1,3},{1,1,1}}; static bb[N]={2,7,4};void main() {int i,j;double a[N+1][N+1],b[N+1];void solve(double a[][N+1],double b[N+1]); int LU(double a[][N+1]); for(i=1;i<=N;i++) { for(j=1;j<=N;j++) a[i][j]=aa[i-1][j-1]; b[i]=bb[i-1]; }if(LU(a)==1){printf("矩阵L如下\n");for(i=1;i<=N;i++){for(j=1;j<=i;j++)if(i==j)printf("%12d ",1);else printf("%12f",a[i][j]);printf("\n");}printf("\n矩阵U如下\n");for(i=1;i<=N;i++){for(j=1;j<=N;j++)if(i<=j)printf("%12f",a[i][j]);else printf("%12s","");printf("\n");}solve(a,b);for(i=1;i<=N;i++)printf("x[%d]=%f ",i,b[i]);printf("\n");}else printf("\nthe LU method failed!\n");getch();}int LU(double a[][N+1])//对N阶矩A阵进行LU分解，结果L、U存放在A的相应位置{int i,j,k,s;double m,n;for(i=2;i<=N;i++)a[i][1]/=a[1][1];for(k=2;k<=N;k++){for(j=k;j<=N;j++){m=0;for(s=1;s<k;s++)m+=a[k][s]*a[s][j];a[k][j]-=m;}if(a[k][k]==0){printf("a[%d][%d]=%d ",k,k,0);return -1;//存在元素akk为0}for(i=k+1;i<=N;i++){n=0;for(s=1;s<k;s++)n+=a[i][s]*a[s][k];a[i][k]=(a[i][k]-n)/a[k][k];}}return 1;//正常结束}void solve(double a[][N+1],double b[N+1])//利用分解的LU求x//回代求解，L和U元素均在A矩阵相应位置；b存放常数列，返回时存放方程组的解{double y[N+1],F;int i,j;y[1]=b[1];for(i=2;i<=N;i++){F=0.0;for(j=1;j<i;j++)F+=a[i][j]*y[j];y[i]=b[i]-F;}b[N]=y[N]/a[N][N];for(i=N-1;i>0;i--){F=0.0;for(j=N;j>i;j--)F+=a[i][j]*b[j];b[i]=(y[i]-F)/a[i][i];}}运行及结果显示：实验十二：Guass-Sediel 算法命名（源程序.cpp 及工作区.dsw ）：GS问题：利用G －S 法求解方程组 ⎪⎩⎪⎨⎧=+--=-+-=--2.453.82102.7210321321321x x x x x x x x x （精度为41021-⨯）//Guass_sediel.cpp#include "iostream.h"#include "math.h"#include "stdio.h"#include<conio.h>#define N 3 //方程的阶数#define MaxK 100 //最大迭代次数#define EPS 0.5e-4 //精度控制static double aa[N][N]={{10,-1,-2},{-1,10,-2},{-1,-1,5}};static double bb[N]={7.2,8.3,4.2};void main(){int i,j;double x[N+1];double a[N+1][N+1],b[N+1];int GaussSeidel(double a[][N+1],double b[],double x[]);for(i=1;i<=N;i++){for(j=1;j<=N;j++)a[i][j]=aa[i-1][j-1];b[i]=bb[i-1];x[i]=0;}if(GaussSeidel(a,b,x)==1){printf("the roots is:");for(i=1;i<=N;i++)printf(" x[%d]=%f ",i,x[i]);printf("\n");}else printf("\nthe G-S method failed!\n");getch();}int GaussSeidel(double a[][N+1],double b[],double x[])//a传入系数矩阵(有效元素为a[1,1]...a[n,n]),b为方程组右边常数列，x传入迭代初值并返回解向量//x**(k+#x)=Gx**(k)+f{int k=1,i,j;double m,max,t[N+1];while(true){printf("k=%d:",k);max=0.0;for(i=1;i<=N;i++){if(a[i][i]==0)return -1;//存在元素akk为0m=0.0;t[i]=x[i];for(j=1;j<=N;j++)if(j!=i)m+=a[i][j]*x[j];x[i]=(b[i]-m)/a[i][i];if(max<fabs(x[i]-t[i]))max=fabs(x[i]-t[i]);printf(" x[%d]=%f ",i,x[i]);}printf("\n");if(max<EPS)return 1;//正常结束elsek++;if(k>MaxK)return -2;//迭代次数越界}}运行及结果显示：实验十三：SOR 算法命名（源程序.cpp 及工作区.dsw ）：sor问题：利用SOR 法求解方程组⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡=⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡----111141111411114111144321x x x x （精度为51021-⨯） //SOR.cpp#include "math.h"#include "stdio.h"#include "conio.h"#define N 4#define MaxK 1000#define EPS 0.5e-5static double aa[N][N]={{-4,1,1,1},{1,-4,1,1},{1,1,-4,1},{1,1,1,-4}}; static double bb[N]={1,1,1,1};void main(){int i,j;double x[N+1];double a[N+1][N+1],b[N+1];int Sor(double a[][N+1],double b[],double w,double x[]);for(i=1;i<=N;i++){for(j=1;j<=N;j++)a[i][j]=aa[i-1][j-1];b[i]=bb[i-1];x[i]=0;}if(Sor(a,b,1.3,x)==1){printf("the roots is:");for(i=1;i<=N;i++)printf(" x[%d]=%f ",i,x[i]);printf("\n");}else printf("\nthe G-S method failed!\n");getch();}int Sor(double a[][N+1],double b[],double w,double x[]) {int i,j,k;double loft,comb;for(k=0;k<=N;k++)x[k]=0;for(i=1;i<=MaxK && fabs(comb)>EPS;i++){printf("k=%d:\t",i);comb=0;for(k=1;k<=N;k++){loft=b[k];for(j=1;j<N+1;j++)loft-=a[k][j]*x[j];if(a[k][k]==0)return -1;else loft=w*loft/a[k][k];x[k]=x[k]+loft;printf("x[%d]=%10f \t",k,x[k]);if(fabs(loft)>comb)comb=fabs(loft);}printf("\n");}if(i>MaxK)return -2;elsereturn 1;}运行结果（w=1）=。