我编写了一个程序，它通过给出一些坐标来计算抛物线的方程，但它只适用于偶数个坐标。如果我输入一个不均匀的数字，它将显示一些无意义的bs。
以下是代码（由于某些格式问题，我无法在此处复制代码）：
#include<iostream>
#include<iomanip>
#include<cmath>
using namespace std;
int main()
{
int i, j, k, n, N;
n = 2;
cout << "Number of data pairs:" << endl;
cin >> N;
double * x = new double[N];
double * y = new double[N];
cout << endl << "Enter the x-axis values:" << endl;
for (i = 0; i<N; i++)
cin >> x[i];
cout << endl << "Enter the y-axis values:" << endl;
for (i = 0; i<N; i++)
cin >> y[i];
double X[5];
for (i = 0; i<2 * n + 1; i++)
{
X[i] = 0;
for (j = 0; j<N; j++)
X[i] = X[i] + pow(x[j], i);
}
double B[3][4], a[3]; //B is the Normal matrix(augmented) that will store the equations, 'a' is for value of the final coefficients
for (i = 0; i <= n; i++)
for (j = 0; j <= n; j++)
B[i][j] = X[i + j];
double Y[3];
for (i = 0; i<n + 1; i++)
{
Y[i] = 0;
for (j = 0; j<N; j++)
Y[i] = Y[i] + pow(x[j], i)*y[j];
}
for (i = 0; i <= n; i++)
B[i][n + 1] = Y[i];
n = n + 1;
for (i = 0; i < n; i++)
{
for (k = i + 1; k < n; k++)
if (B[i][i] < B[k][i])
for (j = 0; j <= n; j++)
{
double temp = B[i][j];
B[i][j] = B[k][j];
B[k][j] = temp;
}
}
for (i = 0; i<n - 1; i++) //loop to perform the gauss elimination
for (k = i + 1; k<n; k++)
{
double t = B[k][i] / B[i][i];
for (j = 0; j <= n; j++)
B[k][j] = B[k][j] - t*B[i][j]; //make the elements below the pivot elements equal to zero or elimnate the variables
}
for (i = n - 1; i >= 0; i--) //back-substitution
{
a[i] = B[i][n];
for (j = 0; j < n; j++)
if (j != i)
a[i] = a[i] - B[i][j] * a[j];
a[i] = a[i] / B[i][i];
}
cout << endl << "The equation is the following:" << endl;;
cout << a[2] << "x^2 + " << a[1] << "x + " << a[0];
cout << endl;
delete[]x;
delete[]y;
system("pause");
return 0;
}
我从3和4坐标得到的输出：
3坐标：
数据点数：
3
输入x轴值：
1
2
3
输入y轴值
1
4
9
等式如下：
1.02762e + 47x ^ 2 + -4.316e + 47x + 3.90495e + 47
4坐标：
数据点数：
4
输入x轴值：
1
2
3
4
输入y轴值
1
4
9
16
等式如下：
1x ^ 2 + -0x + 0
任何想法或提示？
提前致谢
答案 0 :(得分：4)
已发布的评论指出，某些索引超出范围（超出矩阵的大小）。我更喜欢将x，y数据输入为每行一对值，x值和ay值，这是一个很容易的变化（cin＆gt;＆gt; x [i]＆gt;＆gt; y [i] ），但以下示例使用原始问题中的序列。
二次方程拟合的传统方法的示例代码：
#include<iostream>
#include<iomanip>
#include<cmath>
using namespace std;
int main()
{
int i, j, k, N;
cout << "Number of data pairs:" << endl;
cin >> N;
double * x = new double[N];
double * y = new double[N];
cout << endl << "Enter the x-axis values:" << endl;
for (i = 0; i<N; i++)
cin >> x[i];
cout << endl << "Enter the y-axis values:" << endl;
for (i = 0; i<N; i++)
cin >> y[i];
double B[3][4] = {0.0}; // generate augmented matrix
for(k = 0; k < 3; k++){
for (i = 0; i < N; i++) {
for (j = 0; j < 3; j++) {
B[k][j] += pow(x[i], j + k);}
B[k][3] += y[i]*pow(x[i], k);}}
for(k = 0; k < 3; k++){ // invert matrix
double q = B[k][k]; // divide row by B[k][k]
for(i = 0; i < 4; i++){
B[k][i] /= q;}
for(j = 0; j < 3; j++){ // zero out column B[][k]
if(j == k)
continue;
double m = B[j][k];
for(i = 0; i < 4; i++){
B[j][i] -= m*B[k][i];}}}
cout << endl << "The equation is the following:" << endl;;
cout << B[2][3] << " x^2 + " << B[1][3] << " x + " << B[0][3];
cout << endl;
delete[]x;
delete[]y;
system("pause");
return 0;
}
通用度方程的常规代码示例：
#include<iostream>
#include<iomanip>
#include<cmath>
using namespace std;
int main()
{
int d, i, j, k, n;
cout << "Degree of equation:" << endl;
cin >> d;
double **A = new double *[d+1];
for (k = 0; k < d+1; k++) {
A[k] = new double[d+2];
for (i = 0; i < d+2; i++) {
A[k][i] = 0.0;}}
cout << "Number of data pairs:" << endl;
cin >> n;
double * x = new double[n];
double * y = new double[n];
cout << endl << "Enter the x-axis values:" << endl;
for (i = 0; i < n; i++)
cin >> x[i];
cout << endl << "Enter the y-axis values:" << endl;
for (i = 0; i < n; i++)
cin >> y[i];
for(k = 0; k < d+1; k++){
for (i = 0; i < n; i++) {
for (j = 0; j < d+1; j++) {
A[k][j] += pow(x[i], j + k);}
A[k][d+1] += y[i]*pow(x[i], k);}}
for(k = 0; k < d+1; k++){ // invert matrix
double q = A[k][k]; // divide A[k][] by A[k][k]
// if q == 0, would need to swap rows
for(i = 0; i < d+2; i++){
A[k][i] /= q;}
for(j = 0; j < d+1; j++){ // zero out column A[][k]
if(j == k)
continue;
double m = A[j][k];
for(i = 0; i < d+2; i++){
A[j][i] -= m*A[k][i];}}}
cout << endl << "The equation is the following:" << endl;
for(k = d; k >= 2; k--)
cout << A[k][d+1] << " x^" << k << " + ";
cout << A[1][d+1] << " x" << " + " << A[0][d+1] << endl;
for (k = 0; k < d+1; k++)
delete[] A[k];
delete[]A;
delete[]y;
delete[]x;
system("pause");
return 0;
}
示例测试输入：
3
4
1
2
3
4
10
49
142
313
我通常使用一种替代算法，避免必须反转矩阵，这对于高阶多项式更好，但对于二次方程则不需要。
如果对替代算法感兴趣，这里是指向描述算法的pdf文件的链接，并包含伪代码。我有旧的工作代码，但它需要转换（它使用很少支持的cgets（））。
http://rcgldr.net/misc/opls.pdf
示例代码。它真的很古老，最初是在Atari ST上运行的。我将其转换为使用Visual Studio。所有静态的原因是减少了链接器必须处理的符号数。
/*------------------------------------------------------*/
/* fit4.c polynomial fit program */
/* originally written in 1990, minimal updates */
/*------------------------------------------------------*/
/* disable Visual Studio warnings for old functions */
#define _CRT_SECURE_NO_WARNINGS 1
#include <stdio.h>
/* mmax = max # coefficients */
/* nmax = max # points */
/* bfrsz = bfr size */
/* linsz = size of line bfr */
#define mmax 11
#define nmax 300
#define bfrsz 0x2000
#define linsz 64
static void polyf();
static void gcoef();
static void gvar();
static void calc();
static void calcx();
static void rdata();
static void gdata();
static int gtlin();
static char gtchr();
static int conrs();
static char cbfr[64]; /* console response bfr */
static char line[linsz];
static int lineno;
static char sbfr[bfrsz]; /* file params */
static FILE *sfp;
static char *sptr, *send;
static int gteof;
static int wf;
static int m, n; /* input values */
static double x[nmax];
static double y[nmax];
static double w[nmax];
static double b[mmax]; /* generated values */
static double A[mmax];
static double B[mmax];
static double L[mmax];
static double W[mmax];
static double p2[nmax];
static double p1[nmax];
static double p0[nmax];
static double c[mmax]; /* coefficients for y(x) */
static double z[nmax]; /* calculated y[] */
static double xi, zi, vr;
static double D0, D1; /* constants */
static double *pp2; /* pointers to logical p2, p1, p0 */
static double *pp1;
static double *pp0;
static double *ppx;
static double *px, *pf, *pw; /* for gdata */
main()
{
int i;
name0:
printf("\nEnter name of data file: "); /* get file name */
if(!conrs())
return(0);
sfp = fopen(&cbfr[2], "rb"); /* open file */
if(sfp == (FILE *)0){
printf("\nfile not found");
goto name0;}
wf = 0; /* ask for weighting */
printf("\nUsing weights (Y/N)? ");
if(!conrs())
return(0);
if('Y' == (cbfr[2]&0x5f))
wf = 1;
deg0:
printf("\nEnter degree of equation (1-n): "); /* get # terms */
if(!conrs())
return(0);
sscanf(&cbfr[2], "%d", &m);
if(m >= mmax)
goto deg0;
sptr = send = (char *)0;
gteof = 0;
lineno = 0;
gdata(); /* get data */
printf("\n%5d points found ", n);
polyf(); /* generate b[], A[], B[] */
gcoef(); /* generate coefficients */
gvar(); /* generate variance */
for(i = 0; i <= m; i++){
printf("\n%3d b%12.4le A%12.4le B%12.4le",
i, b[i], A[i], B[i]);
printf(" L%12.4le W%12.4le", L[i], W[i]);}
printf("\nvariance = %12.4le\n", vr);
for(i = m; i; i--)
printf("%12.4le X**%1d + ", c[i], i);
printf("%12.4le\n", c[0]);
for(i = 0; i < n; i++){ /* calculate results */
xi = x[i];
calc();
z[i] = zi;}
for(i = 0; i < n; i += 1) /* display results */
printf("\n%14.6le %14.6le %14.6le %14.6le",
x[i], y[i], z[i], y[i]-z[i]);
printf("\n");
return(0);
}
/*------------------------------------------------------*/
/* polyf poly fit */
/* in: x[], y[], w[], n, m */
/* out: b[], A[], B[], L[], W[] */
/*------------------------------------------------------*/
static void polyf()
{
int i, j;
D0 = (double)0.; /* init */
D1 = (double)1.;
pp2 = p2;
pp1 = p1;
pp0 = p0;
j = 0;
A[j] = D0; /* calc A, p[j], p[j-1], L, W */
L[j] = D0; /* note A[0] not used */
W[j] = D0;
for(i = 0; i < n; i++){
pp0[i] = D1;
pp1[i] = D0;
L[j] += w[i];
W[j] += w[i]*y[i];}
B[0] = D0;
b[j] = W[j]/L[j];
for(j = 1; j <= m; j++){
ppx = pp2; /* save old p[j], p[j-1] */
pp2 = pp1;
pp1 = pp0;
pp0 = ppx;
A[j] = D0; /* calc A */
for(i = 0; i < n; i++){
A[j] += w[i]*x[i]*pp1[i]*pp1[i]/L[j-1];}
L[j] = D0; /* calc p[j], L, W */
W[j] = D0;
for(i = 0; i < n; i++){
pp0[i] = (x[i]-A[j])*pp1[i]-B[j-1]*pp2[i];
L[j] += w[i]*pp0[i]*pp0[i];
W[j] += w[i]*y[i]*pp0[i];}
B[j] = L[j]/L[j-1]; /* calc B[], b[] */
b[j] = W[j]/L[j];}
}
/*------------------------------------------------------*/
/* gcoef generate coefficients */
/* in: b[], A[], B[] */
/* out: c[] */
/* uses: p0[], p1[], p2[] */
/*------------------------------------------------------*/
static void gcoef()
{
int i, j;
for(i = 0; i <= m; i++){ /* init */
c[i] = p2[i] = p1[i] = p0[i] = 0.;}
p0[0] = D1;
c[0] += b[0]*p0[0];
for(j = 1; j <= m; j++){ /* generate coefs */
p2[0] = p1[0];
p1[0] = p0[0];
p0[0] = -A[j]*p1[0]-B[j-1]*p2[0];
c[0] += b[j]*p0[0];
for(i = 1; i <= j; i++){
p2[i] = p1[i];
p1[i] = p0[i];
p0[i] = p1[i-1]-A[j]*p1[i]-B[j-1]*p2[i];
c[i] += b[j]*p0[i];}}
}
/*------------------------------------------------------*/
/* gvar generate variance */
/*------------------------------------------------------*/
static void gvar()
{
int i;
double tt;
vr = 0.;
for(i = 0; i < n; i++){
xi = x[i];
calc();
tt = y[i]-zi;
vr += tt*tt;}
vr /= n-m-1;
}
/*------------------------------------------------------*/
/* calc calc zi, given xi */
/* in: c[] */
/*------------------------------------------------------*/
static void calc ()
{
int i;
zi = c[m];
for(i = m-1; i >= 0; i--)
zi = zi*xi + c[i];
}
/*------------------------------------------------------*/
/* calcx calc zi, given xi */
/* in: b[], A[], B[] */
/*------------------------------------------------------*/
static void calcx()
{
int i;
double q2, q1, q0;
if(m == 0){
zi = b[0];
return;}
if(m == 1){
zi = b[0]+(xi-A[1])*b[1];
return;}
q1 = b[m];
q0 = b[m-1]+(xi-A[m])*q1;
for(i = m-2; i >= 0; i--){
q2 = q1;
q1 = q0;
q0 = b[i]+(xi-A[i+1])*q1-B[i+1]*q2;}
zi = q0;
}
/*------------------------------------------------------*/
/* gdata get data */
/*------------------------------------------------------*/
static void gdata()
{
px = &x[0];
pf = &y[0];
pw = &w[0];
while(1){
gtlin(); /* get a line */
if(gteof)
break;
if(lineno == nmax){
printf("\ntoo many points\n");
break;}
if(wf){ /* stuff values */
sscanf(line, "%le%le%le", px, pf, pw);}
else{
sscanf(line, "%le%le", px, pf);
*pw = 1.0;}
px++; /* bump ptrs */
pf++;
pw++;}
fclose(sfp); /* close file */
n = lineno; /* set # points */
}
/*------------------------------------------------------*/
/* gtlin get a line of data */
/*------------------------------------------------------*/
static int gtlin()
{
char chr;
int col;
col = 0;
while(1){
chr = gtchr();
switch(chr){
case 0x0a: /* line feed */
lineno++;
return(col);
case 0x1a:
return(col);
default:
line[col] = chr;
col++;
if(col >= linsz){
printf("line # %d too long\n%s",lineno, line);
return(col);}}}
}
/*------------------------------------------------------*/
/* gtchr get a char */
/*------------------------------------------------------*/
static char gtchr()
{
int cnt;
if(gteof) /* check for eof */
return(0x1a);
if(sptr == send){
if(!(cnt = (int) fread(sbfr, 1, bfrsz, sfp))){
fclose(sfp);
gteof = 1;
return(0x1a);}
sptr = sbfr;
send = sbfr+cnt;}
return(*sptr++);
}
/*------------------------------------------------------*/
/* conrs get string from console */
/*------------------------------------------------------*/
static int conrs()
{
int i;
memset(cbfr, 0, sizeof(cbfr)); /* get a line */
cbfr[0] = sizeof(cbfr)-2;
fgets(cbfr+2, sizeof(cbfr)-2, stdin);
cbfr[1] = (char)(strlen(&cbfr[2])-1);
i = cbfr[1];
cbfr[2+i] = 0;
return(i);
}
示例数据文件。我把它命名为fitdat.txt：
1 1
2 4
3 9
编译并运行程序。输入数据文件的名称，然后在提示使用权重时输入N，然后输入2表示要生成的公式的程度。 （如果使用权重，第一列将是加权因子，权重为2将具有相同数据点的两个实例相同，但权重可以是1.5之类的值。）
答案 1 :(得分：1)
您正在阅读未初始化的值a[j]：
for (i = n - 1; i >= 0; i--) //back-substitution
{
a[i] = B[i][n];
for (j = 0; j < n; j++)
if (j != i)
a[i] = a[i] - B[i][j] * a[j];
a[i] = a[i] / B[i][i];
}