#include <iostream.h>
#include <iomanip.h>
#include <fstream.h>
#include <time.h>
#include "bignumber.h"

int f(__int64 *a,int b,int a1)
//求f(x)=arctan(1/b)的函數，b>1,否則收斂太慢甚至不收斂
//本函數要求b^2在655535以內，即unsigned int數據類型的范圍內，
//如需支持更大的數，只需調整相關函數的數值類型，由于本實驗沒有用到
//所以為提高程序速度，在夠用的情況下采用盡可能小的數據類型。

//arctanx=1-x^3/3+x^5/5-x^7/7+…
{
	int i,j,t;
	__int64 *c=new __int64[a1];
	for (i=0;i<a1;i++)
		a[i]=c[i]=0;
	c[a1-1]=1;
	unsigned int b2=b*b;
	divide (c,b,a1);
	plus(a,c,a1);
	for (i=1;;i++){
		multiply(c,2*i-1,a1);
		divide(c,2*i+1,a1);
		divide(c,b2,a1);
		if(i%2==0)plus(a,c,a1);
		else minus(a,c,a1);
		t=0;
		for (j=0;j<a1;j++)
			if (c[j]!=0){
				t=1;
				break;
			}
		if(t==0)break;
	}
	delete c;
	return i;
}

main()
{
	int n;
	cout<<"輸入所需精度!";
	cin>>n;
	clock_t p=clock();
	int m=n/9+3;
	__int64 *a=new __int64[m];
	__int64 *b=new __int64[m];
	int i;
	ofstream file;
	file.open("1.txt");
	i=f(a,5,m);
	cout<<"Arctan(1/5)迭代次數:"<<i<<endl;
	file<<"Arctan(1/5)迭代次數:"<<i<<endl;
	i=f(b,239,m);
	cout<<"Arctan(1/239)迭代次數:"<<i<<endl;
	file<<"Arctan(1/239)迭代次數:"<<i<<endl;	
	multiply(a,16,m);
	multiply(b,4,m);
	minus(a,b,m);
	double q=(clock()-p)/(double)CLOCKS_PER_SEC;
	cout<<"運算時間:"<<q<<"秒"<<endl;
	file<<"運算時間:"<<q<<"秒"<<endl;
	cout<<(long)a[m-1]<<'.';
	file<<(long)a[m-1]<<'.';
	for (i=m-2;i>=2;i--){
		cout<<setw(9)<<setfill('0')<<(long)a[i]<<' ';
		file<<setw(9)<<setfill('0')<<(long)a[i]<<' ';
	}
	file.close ();
	return 0;
}