是否有一个C++库可以计算大数的n次根(即无法放入unsigned long long
的数字)?
http://en.wikipedia.org/wiki/Nth_root
迭代算法非常简单:x(k+1) = [(n - 1) * x(k) + A / x(k)^(n - 1)] / n
我猜这取决于你想要比2^64大多少。仅使用双精度浮点数可达到约10^-9的精度。我用C语言编写了一个测试程序:
#include <stdio.h>
#include <math.h>
int main(int argc, char **argv)
{
unsigned long long x;
double dx;
int i;
//make x the max possible value
x = ~0ULL;
dx = (double)x;
printf("Starting with dx = %f\n", dx);
//print the 2th to 20th roots
for (i = 2; i < 21; i++)
{
printf("%dth root %.15f\n", i, pow(dx, 1.0/i));
}
return 0;
}
生成以下输出:
Starting with dx = 18446744073709551616.000000
2th root 4294967296.000000000000000
3th root 2642245.949629130773246
4th root 65536.000000000000000
5th root 7131.550214521852467
6th root 1625.498677215435691
7th root 565.293831000991759
8th root 256.000000000000000
9th root 138.247646578215154
10th root 84.448506289465257
11th root 56.421840319745364
12th root 40.317473596635935
13th root 30.338480458853493
14th root 23.775908626191171
15th root 19.248400577313866
16th root 16.000000000000000
17th root 13.592188707483222
18th root 11.757875938204789
19th root 10.327513583579238
20th root 9.189586839976281
然后我将每个根与Wolfram Alpha进行比较,以获得我上面引用的误差。
根据您的应用程序,也许这已经足够好了。
~
运算符,即x = ~0ULL
。 - MSalters // n1 = <input>, n2 = <base limit>, nmrk = <number of cycles>
long double n3 = 0, n2 = 0, n1 = input_number, n4 = 0;
long double mk = 0, tptm = 0, mnk = 0, dad = 0;
for (n3 = 0; tptm != n1 || mnk > 65535 ; nmrk++) {
n4 += 0.19625;
n2 += 0.15625;
n3 += 0.015625;
mk += 0.0073125;
dad += 0.00390625;
mnk = pow(n1, 1.0/(n4+n2+mk+n3+dad));
tptm = pow((mnk), (n4+n2+mk+n3+dad));
}
if (tptm - n1 < 1)
{
uint64_t y = (tptm);
return make_tuple(nmrk, (n1 - y), mnk);
}
我发现这个比之前快了几分钟
长除法是计算任何正实数的第n个根的最佳方法。它提供了每个计算数字的最佳精度。不需要初始猜测和迭代逼近。