用三角函数做最佳平方逼近（C++实现）

用三角函数做最佳平方逼近（C++实现）

#include<iostream>
#include<cstdio> 
#include<cmath>
#define pi  3.1415926535
using namespace std;
double lim(int num,double x[],double y[],double t,int aa){
	//这里判断num和aa为奇数还是偶数；
	int a[2];
	int a1=num/2;
	int a2=aa/2;
	if(a1<(num/2))
	a[0]=0;//0代表奇数；
	else
	a[0]=1; 
	if(a2<(aa/2))
	a[1]=0;//0代表奇数； 
	else
	a[1]=1; 
	if(a[1]==0){
	 double ak[a2+1];
	 double bk[a2];
	 for (int i=0;i<a2+1;i++){
	 	 double sum1=0;
	 	 for (int j=0;j<num;j++){
	 	   sum1=sum1+y[j]*cos(2*pi*j*i/(num));
		  }
		  ak[i]=sum1*(2/num);
	 }
	for (int i=0;i<a2;i++){
		 double sum1=0;
		 for (int j=0;j<num;j++){
		 	 sum1=sum1+y[j]*sin(2*pi*j*(i+1)/(num));
		 }
		 bk[i]=sum1*(2/num);
	 }	
	 double s1=ak[0]/2;
	   for(int i=1;i<a2+1;i++){
	   	s1=s1+ak[i]*cos(i*aa);
	   }
	   double s2=0;
	   for(int i=0;i<a2;i++){
	   	s2=s2+bk[i]*cos((i+1)*aa);
	   }
	   cout<<"输出系数a表格"<<"\n";
	   for (int i=0;i<a2+1;i++){
	   	cout<<ak[i]<<"  ";
	   } 
	   cout<<"\n\n";
	   cout<<"输出系数b表格"<<"\n";
	   for (int i=0;i<a2;i++){
	   	cout<<bk[i]<<"  ";
	   } 
	   cout<<"\n\n";
	   printf("输出计算的值:");
	 	cout<<s1+s2;
	}else{
	double ak[a2+1];
	double bk[a2-1];
	for (int i=0;i<a2+1;i++){
		double sum1=0;
		for (int j=0;j<num;j++){
			sum1=sum1+y[j]*cos(2*pi*j*i/(num));
		}
		ak[i]=sum1*(2/num);
	}
	for (int i=0;i<a2-1;i++){
		 double sum1=0;
		 for (int j=0;j<num;j++){
		 	 sum1=sum1+y[j]*sin(2*pi*j*(i+1)/(num));
		 }
		 bk[i]=sum1*(2/num);
	 }	
	 ///////
	 double s1=ak[0]/2;
	   for(int i=1;i<a2+1;i++){
	   	s1=s1+ak[i]*cos(i*aa);
	   }
	   double s2=0;
	   for(int i=0;i<a2-1;i++){
	   	s2=s2+bk[i]*cos((i+1)*aa);
	   }
	   cout<<"输出系数a表格"<<"\n";
	   for (int i=0;i<a2+1;i++){
	   	cout<<ak[i]<<"  ";
	   } 
	   cout<<"\n\n";
	   cout<<"输出系数b表格"<<"\n";
	   for (int i=0;i<a2-1;i++){
	   	cout<<bk[i]<<"  ";
	   } 
	   cout<<"\n\n";
	   printf("输出计算的值:");
	 	cout<<s1+s2; 			
	}	 
}
int main(){
    int num;
    cout<<"输入节点的个数："; 
    scanf("%d",&num);
    double x[num],y[num];
    for (int i=0;i<num;i++){
    	cout<<"输入第"<<i+1<<"个x的值："; 
	    scanf("%lf",&x[i]);
		cout<<"输入第"<<i+1<<"个y的值：";
		scanf("%lf",&y[i]);
	}
	cout<<"输入要计算的x值：";
	double t;
	cin>>t; 
	cout<<"输入做多少项的三角多项式拟合"; 
	int aa;
	cin>>aa;
	double tt[num];
	for(int i=0;i<num;i++){
		tt[i]=((2*x[i]-(x[0]+x[num-1]))/(x[0]+x[num-1]))*pi;
	}
	lim(num,tt,y,t,aa);//lim做最佳三角平方逼近 
	return 0;  
}