首页 / C++ / 多项式正反算解算源码 自己写的 C++
多项式正反算解算源码 自己写的 C++
内容导读
互联网集市收集整理的这篇技术教程文章主要介绍了多项式正反算解算源码 自己写的 C++，小编现在分享给大家，供广大互联网技能从业者学习和参考。文章包含8290字，纯文字阅读大概需要12分钟。
内容图文
一次，二次、三次多项式系数解算是工程项目里最常用到的方法，就是常用的数学公式，如果从头写还是比较费劲，涉及到高斯解方程等，这里将源码分享给大家，已经经过大量的测试，解算算法是完全正确的。直接复制即可使用，如果能帮到大家，希望能点赞。
//solve one order polynomial
void PolynCal1(double *x1, double *y1, double *x2,
double *y2, int n, double *cofx, double *cofy)
{
for(int i=0; i<n; i++)
{
x1[i]=(cofx[0]+cofx[1]*x2[i]+cofx[2]*y2[i]);
y1[i]=(cofy[0]+cofy[1]*x2[i]+cofy[2]*y2[i]);
}
}
void PolynCoff3(double *x1, double *y1, double *x2,
double *y2, int n, double *cofx, double *cofy,
double *rmsx, double *rmsy)
{
double a[10],aa[100],al[10];
int i;
//x
memset(aa,0,sizeof(double)*100);
memset(al,0,sizeof(double)*10);
for(i=0; i<n; i++)
{
a[0] = 1.0;
a[1] = x2[i]; a[2] = y2[i];
a[3] = x2[i]*x2[i]; a[4] = x2[i]*y2[i]; a[5] = y2[i]*y2[i];
a[6] = x2[i]*x2[i]*x2[i]; a[7] = x2[i]*x2[i]*y2[i];
a[8] = x2[i]*y2[i]*y2[i]; a[9] = y2[i]*y2[i]*y2[i];
EquationNorm(a,10,x1[i],aa,al,1.0);
}
Agaus(aa, al, 10);
memcpy(cofx, al, sizeof(double)*10);
//y
memset(aa,0,sizeof(double)*100);
memset(al,0,sizeof(double)*10);
for(i=0; i<n; i++)
{
a[0] = 1.0;
a[1] = x2[i]; a[2] = y2[i];
a[3] = x2[i]*x2[i]; a[4] = x2[i]*y2[i]; a[5] = y2[i]*y2[i];
a[6] = x2[i]*x2[i]*x2[i]; a[7] = x2[i]*x2[i]*y2[i];
a[8] = x2[i]*y2[i]*y2[i]; a[9] = y2[i]*y2[i]*y2[i];
EquationNorm(a,10,y1[i],aa,al,1.0);
}
Agaus(aa, al, 10);
memcpy(cofy, al, sizeof(double)*10);
if(rmsx)
{
for(i=0; i<n; i++)
{
rmsx[i]=(cofx[0]+
cofx[1]*x2[i]+cofx[2]*y2[i]+
cofx[3]*x2[i]*x2[i]+cofx[4]*x2[i]*y2[i]+
cofx[5]*y2[i]*y2[i] + cofx[6]*x2[i]*x2[i]*x2[i] + cofx[7]*x2[i]*x2[i]*y2[i]+
cofx[8]*x2[i]*y2[i]*y2[i] + cofx[9]*y2[i]*y2[i]*y2[i])-x1[i];
}
}
if(rmsy)
{
for(i=0; i<n; i++)
{
rmsy[i]=(cofy[0]+
cofy[1]*x2[i]+cofy[2]*y2[i]+
cofy[3]*x2[i]*x2[i]+cofy[4]*x2[i]*y2[i]+
cofy[5]*y2[i]*y2[i] + cofy[6]*x2[i]*x2[i]*x2[i] + cofy[7]*x2[i]*x2[i]*y2[i]+
cofy[8]*x2[i]*y2[i]*y2[i] + cofy[9]*y2[i]*y2[i]*y2[i])-y1[i];
}
}
}
//solve one order polynomial
void PolynCoff2(double *x1, double *y1, double *x2,
double *y2, int n, double *cofx, double *cofy,
double *rmsx, double *rmsy)
{
double a[6],aa[36],al[6];
int i;
//x
memset(aa,0,sizeof(double)*36);
memset(al,0,sizeof(double)*6);
for(i=0; i<n; i++)
{
a[0] = 1.0;
a[1] = x2[i]; a[2] = y2[i];
a[3] = x2[i]*x2[i]; a[4] = x2[i]*y2[i]; a[5] = y2[i]*y2[i];
EquationNorm(a,6,x1[i],aa,al,1.0);
}
Agaus(aa, al, 6);
memcpy(cofx, al, sizeof(double)*6);
//y
memset(aa,0,sizeof(double)*36);
memset(al,0,sizeof(double)*6);
for(i=0; i<n; i++)
{
a[0] = 1.0;
a[1] = x2[i]; a[2] = y2[i];
a[3] = x2[i]*x2[i]; a[4] = x2[i]*y2[i]; a[5] = y2[i]*y2[i];
EquationNorm(a,6,y1[i],aa,al,1.0);
}
Agaus(aa, al, 6);
memcpy(cofy, al, sizeof(double)*6);
if(rmsx)
{
for(i=0; i<n; i++)
{
rmsx[i]=(cofx[0]+
cofx[1]*x2[i]+cofx[2]*y2[i]+
cofx[3]*x2[i]*x2[i]+cofx[4]*x2[i]*y2[i]+
cofx[5]*y2[i]*y2[i])-x1[i];
}
}
if(rmsy)
{
for(i=0; i<n; i++)
{
rmsy[i]=(cofy[0]+
cofy[1]*x2[i]+cofy[2]*y2[i]+
cofy[3]*x2[i]*x2[i]+cofy[4]*x2[i]*y2[i]+
cofy[5]*y2[i]*y2[i])-y1[i];
}
}
}
void PolynCal2(double *x1, double *y1, double *x2,
double *y2, int n, const double *cofx, const double *cofy)
{
for(int i=0; i<n; i++)
{
x1[i]=(cofx[0]+
cofx[1]*x2[i]+cofx[2]*y2[i]+
cofx[3]*x2[i]*x2[i]+cofx[4]*x2[i]*y2[i]+cofx[5]*y2[i]*y2[i]);
y1[i]=(cofy[0]+
cofy[1]*x2[i]+cofy[2]*y2[i]+
cofy[3]*x2[i]*x2[i]+cofy[4]*x2[i]*y2[i]+cofy[5]*y2[i]*y2[i]);
}
}
void PolynCal3(double *x1, double *y1, double *x2,
double *y2, int n, const double *cofx, const double *cofy)
{
for(int i=0; i<n; i++)
{
x1[i]=(cofx[0]+
cofx[1]*x2[i]+cofx[2]*y2[i]+
cofx[3]*x2[i]*x2[i]+cofx[4]*x2[i]*y2[i]+
cofx[5]*y2[i]*y2[i] + cofx[6]*x2[i]*x2[i]*x2[i] + cofx[7]*x2[i]*x2[i]*y2[i]+
cofx[8]*x2[i]*y2[i]*y2[i] + cofx[9]*y2[i]*y2[i]*y2[i]);
y1[i]=(cofy[0]+
cofy[1]*x2[i]+cofy[2]*y2[i]+
cofy[3]*x2[i]*x2[i]+cofy[4]*x2[i]*y2[i]+
cofy[5]*y2[i]*y2[i] + cofy[6]*x2[i]*x2[i]*x2[i] + cofy[7]*x2[i]*x2[i]*y2[i]+
cofy[8]*x2[i]*y2[i]*y2[i] + cofy[9]*y2[i]*y2[i]*y2[i]);
}
}
//solve one order polynomial
void PolynCoff1(double *x1, double *y1, double *x2,
double *y2, int n, double *cofx, double *cofy,
double *rmsx, double *rmsy)
{
double a[3],aa[9],al[3];
int i;
//x
memset(aa,0,sizeof(double)*9);
memset(al,0,sizeof(double)*3);
for(i=0; i<n; i++)
{
a[0] = 1.0;
a[1] = x2[i]; a[2] = y2[i];
EquationNorm(a,3,x1[i],aa,al,1.0);
}
Agaus(aa, al, 3);
memcpy(cofx, al, sizeof(double)*3);
//y
memset(aa,0,sizeof(double)*9);
memset(al,0,sizeof(double)*3);
for(i=0; i<n; i++)
{
a[0] = 1.0;
a[1] = x2[i]; a[2] = y2[i];
EquationNorm(a,3,y1[i],aa,al,1.0);
}
Agaus(aa, al, 3);
memcpy(cofy, al, sizeof(double)*3);
//rmsx
if(rmsx)
{
for(i=0; i<n; i++)
rmsx[i]=(cofx[0]+cofx[1]*x2[i]+cofx[2]*y2[i])-x1[i];
}
//rmsy
if(rmsy)
{
for(i=0; i<n; i++)
rmsy[i]=(cofy[0]+cofy[1]*x2[i]+cofy[2]*y2[i])-y1[i];
}
}
void EquationNorm(double *a, int n, double b,
double *aa, double *ab, double p)
{
int i,j;
for (i=0; i<n; i++)
{
for (j=0; j<n; j++)
*aa++ += p* a[i] * a[j];
*ab++ += p* a[i] * b;
}
}
int Agaus(double *a, double *b, int n)
{
int l = 1, k, i, j, is = 0, p, q;
boost::scoped_array<int> js(new int[n]);
double d,t;
for (k=0; k<=n-2; k++) {
d=0.0;
for (i=k; i<=n-1; i++) {
for (j=k; j<=n-1; j++) {
t = std::fabs(a[i*n+j]);
if (t>d) {
d = t;
js[k] = j;
is = i;
}
}
}
if (d+1.0 == 1.0)
l = 0;
else {
if (js[k] != k)
for (i=0; i<=n-1; i++) {
p = i*n+k;
q = i*n+js[k];
t = a[p];
a[p] = a[q];
a[q] = t;
}
if (is != k) {
for (j=k; j<=n-1; j++) {
p = k*n+j;
q = is*n+j;
t = a[p];
a[p] = a[q];
a[q] = t;
}
t = b[k];
b[k] = b[is];
b[is] = t;
}
}
if (l == 0)
return 0;
d = a[k*n+k];
for (j=k+1; j<=n-1; j++) {
p = k*n+j;
a[p] = a[p]/d;
}
b[k] = b[k]/d;
for (i=k+1; i<=n-1; i++) {
for (j=k+1; j<=n-1; j++) {
p = i*n+j;
a[p] = a[p]-a[i*n+k]*a[k*n+j];
}
b[i] = b[i]-a[i*n+k]*b[k];
}
}
d = a[(n-1)*n+n-1];
if (std::fabs(d)+1.0 == 1.0)
return 0;
b[n-1] = b[n-1]/d;
for (i=n-2; i>=0; i--) {
t = 0.0;
for (j=i+1; j<=n-1; j++)
t = t+a[i*n+j]*b[j];
b[i] = b[i]-t;
}
js[n-1] = n-1;
for (k=n-1; k>=0; k--) {
if (js[k] != k) {
t = b[k];
b[k] = b[js[k]];
b[js[k]] = t;
}
}
return 1;
}
内容总结
以上是互联网集市为您收集整理的多项式正反算解算源码 自己写的 C++全部内容，希望文章能够帮你解决多项式正反算解算源码 自己写的 C++所遇到的程序开发问题。 如果觉得互联网集市技术教程内容还不错，欢迎将互联网集市网站推荐给程序员好友。
内容备注
版权声明：本文内容由互联网用户自发贡献，该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务，不拥有所有权，不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容， 请发送邮件至 gblab@vip.qq.com 举报，一经查实，本站将立刻删除。
内容手机端
扫描二维码推送至手机访问。