public static bigNum rational(string inp, int acc)
{
//--------------有理逼近---------------
bigNum.maxlen = acc + 10;
bigNum input = new bigNum(inp);
int loop = acc * 2;
bigNum countDown = new bigNum(loop.ToString());
bigNum countDownPlus = countDown + one;
bigNum powerRoot = new bigNum(((loop + 1) / 2).ToString());
bigNum temp = two;
int r;
for (int i = 1; ; ++i)
{
if (temp > input)
{
r = i - 1;
break;
}
temp = temp * two;
}
temp = temp / two;
//temp.show();
bigNum a = input / temp - one;
//a.show();
bigNum R = new bigNum(r.ToString());
//R.show();
/*
*
* 以上部分已将输入x转化为(1+a)*2^r
*
*/
times[2] = loop;
temp = new bigNum("0");
while (loop-- > 0)
{
//根据有理逼近公式进行迭代
temp = ((powerRoot * powerRoot * a) / (countDownPlus + temp));
countDownPlus = countDown;
countDown = new bigNum(loop.ToString());
powerRoot = new bigNum(((loop + 1) / 2).ToString());
}
return((a / (one + temp)) + (R * ln2));
}
//根据要求精度四舍五入
public static bigNum round(bigNum input, int acc)
{
bigNum ans = new bigNum(input);
if (ans.num[ans.dot - acc - 1] > 5)
{
ans.num[ans.dot - acc]++;
for (int i = ans.dot - acc; i < ans.cnt; ++i)
{
if (ans.num[i] >= bigNum.mod)
{
ans.num[i + 1] += ans.num[i] / bigNum.mod;
ans.num[i] %= bigNum.mod;
if (i == ans.cnt - 1)
{
ans.cnt++;
}
}
}
}
return(ans);
}
public static bigNum romberg(string inp, int acc)
{
//--------------Romberg数值积分---------------
bigNum.maxlen = acc + 10;
bigNum input = new bigNum(inp);
bigNum[] power4 = new bigNum[50];
bigNum[] stack_T = new bigNum[50];
int cnt_x = 1;
bigNum temp = two;
int r;
for (int i = 1; ; ++i)
{
if (temp > input)
{
r = i - 1;
break;
}
temp = temp * two;
}
temp = temp / two;
//temp.show();
bigNum a = input / temp;
//a.show();
bigNum R = new bigNum(r.ToString());
//R.show();
/*
*
* 以上部分已将输入x转化为a*2^r
*
*/
bigNum h = (a - one);
bigNum half_h = h / two;
stack_T[0] = half_h * (one + f(a));
power4[0] = new bigNum(one);
bigNum x;
bigNum delta;
int n;
bool returnflag;
for (n = 1; ; ++n)
{
//MessageBox.Show(n.ToString());
power4[n] = power4[n - 1] * four;
temp = new bigNum("0");
x = one + half_h;
for (int i = 0; i < cnt_x; ++i)
{
temp = temp + f(x);
x = x + h;
}
h = half_h;
half_h = h / two;
cnt_x <<= 1;
stack_T[n] = (stack_T[n - 1] / two) + (h * temp);//T0(n)
returnflag = true;
for (int i = n - 1; i >= 0; --i)
{
temp = ((power4[n - i] * stack_T[i + 1]) - stack_T[i]) / (power4[n - i] - one);
if (i == 0)
{
delta = temp - stack_T[0];
for (int j = delta.cnt - 1; j >= delta.dot - acc - 1; --j)
{
if (delta.num[j] > 0)
{
returnflag = false;
break;
}
}
if (returnflag)
{
//精度达到要求
times[1] = n;
//MessageBox.Show("romberg迭代到" + n.ToString() + "次");
return(temp + R * ln2);
}
}
stack_T[i] = temp;
}
}
}
public static bigNum halfTaylor(string inp, int acc)
{
//--------------减半Taylor展开---------------
bigNum.maxlen = acc + 10;
bigNum input = new bigNum(inp);
bigNum temp = two;
int r;
for (int i = 1; ; ++i)
{
if (temp > input)
{
r = i - 1;
break;
}
temp = temp * two;
}
temp = temp / two;
//temp.show();
bigNum a = input / temp - one;
//a.show();
bigNum R = new bigNum(r.ToString());
//R.show();
/*
*
* 以上部分已将输入x转化为(1+a)*2^r
*
*/
bigNum ans = new bigNum("0"), an = new bigNum(a);
bigNum delta;
int n = 1;
bigNum offset = new bigNum("0");
bool flag;
do
{
flag = false;
delta = an / (new bigNum(n.ToString()));
for (int i = delta.dot - 1; i >= delta.dot - acc - 1 && i >= 0; --i)
{
if (delta.num[i] != 0)
{
//精度达到要求
flag = true;
break;
}
}
if (flag == false)
{
break;
}
ans = ans + delta;
n++;
an = an * a;
an.neg = (n % 2 == 0 ? true : false);
//递推下一项
} while (flag);
times[0] = n;
//MessageBox.Show("taylor迭代到" + n.ToString() + "次");
ans = ans + ln2 * R;
return(ans);
}
//求倒数
private static bigNum f(bigNum inp)
{
return(one / inp);
}
