212 lines
4.6 KiB
C++
212 lines
4.6 KiB
C++
// ------------------------------------------------------------------
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//
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// Adapted From: http://stackoverflow.com/questions/5905822/function-to-find-the-nth-digit-of-pi
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// Other references:
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// http://bellard.org/pi/pi_n2/pi_n2.html
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// https://web.archive.org/web/20150627225748/http://en.literateprograms.org/Pi_with_the_BBP_formula_%28Python%29
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//
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// ------------------------------------------------------------------
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/*
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* Computation of the n'th decimal digit of \pi with very little memory.
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* Written by Fabrice Bellard on January 8, 1997.
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*
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* We use a slightly modified version of the method described by Simon
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* Plouffe in "On the Computation of the n'th decimal digit of various
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* transcendental numbers" (November 1996). We have modified the algorithm
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* to get a running time of O(n^2) instead of O(n^3log(n)^3).
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*
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* This program uses mostly integer arithmetic. It may be slow on some
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* hardwares where integer multiplications and divisons must be done
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* by software. We have supposed that 'int' has a size of 32 bits. If
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* your compiler supports 'long long' integers of 64 bits, you may use
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* the integer version of 'mul_mod' (see HAS_LONG_LONG).
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*/
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#include "computePi.hpp"
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#include <cstdlib>
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#include <cmath>
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#include <string>
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/* return the inverse of x mod y */
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int inv_mod(int x, int y) {
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int q, u, v, a, c, t;
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u = x;
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v = y;
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c = 1;
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a = 0;
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do {
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q = v / u;
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t = c;
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c = a - q * c;
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a = t;
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t = u;
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u = v - q * u;
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v = t;
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} while (u != 0);
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a = a % y;
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if (a < 0)
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a = y + a;
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return a;
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}
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/* return (a^b) mod m */
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int pow_mod(int a, int b, int m) {
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int r, aa;
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r = 1;
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aa = a;
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while (1) {
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if (b & 1)
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r = mul_mod(r, aa, m);
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b = b >> 1;
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if (b == 0)
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break;
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aa = mul_mod(aa, aa, m);
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}
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return r;
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}
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/* return true if n is prime */
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int is_prime(int n) {
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int r, i;
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if ((n % 2) == 0)
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return 0;
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r = (int)(sqrt(n));
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for (i = 3; i <= r; i += 2)
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if ((n % i) == 0)
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return 0;
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return 1;
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}
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/* return the prime number immediatly after n */
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int next_prime(int n) {
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do {
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n++;
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} while (!is_prime(n));
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return n;
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}
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unsigned int computePiDigit(int n) {
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int av, a, vmax, N, num, den, k, kq, kq2, t, v, s, i;
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double sum = 0;
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N = (int)((n + 20) * std::log(10) / std::log(2));
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for (a = 3; a <= (2 * N); a = next_prime(a)) {
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vmax = (int)(std::log(2 * N) / std::log(a));
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av = 1;
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for (i = 0; i < vmax; i++)
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av = av * a;
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s = 0;
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num = 1;
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den = 1;
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v = 0;
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kq = 1;
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kq2 = 1;
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for (k = 1; k <= N; k++) {
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t = k;
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if (kq >= a) {
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do {
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t = t / a;
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v--;
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} while ((t % a) == 0);
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kq = 0;
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}
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kq++;
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num = mul_mod(num, t, av);
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t = (2 * k - 1);
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if (kq2 >= a) {
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if (kq2 == a) {
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do {
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t = t / a;
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v++;
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} while ((t % a) == 0);
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}
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kq2 -= a;
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}
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den = mul_mod(den, t, av);
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kq2 += 2;
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if (v > 0) {
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t = inv_mod(den, av);
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t = mul_mod(t, num, av);
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t = mul_mod(t, k, av);
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for (i = v; i < vmax; i++)
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t = mul_mod(t, a, av);
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s += t;
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if (s >= av)
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s -= av;
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}
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}
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t = pow_mod(10, n - 1, av);
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s = mul_mod(s, t, av);
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sum = std::fmod(sum + (double)s / (double)av, 1.0);
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}
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return static_cast<unsigned int>(sum * 1e9 / 100000000);
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}
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// ------------------------------------------------------------------
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//
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// Code adapted from this source: https://web.archive.org/web/20150627225748/http://en.literateprograms.org/Pi_with_the_BBP_formula_%28Python%29
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//
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// ------------------------------------------------------------------
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double powneg(unsigned long long b, long long pow) {
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double r = std::pow(b, -pow);
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return 1.0 / r;
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}
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unsigned long long s(unsigned long long j, unsigned long long n) {
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const unsigned long long D = 14;
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const unsigned long long M = static_cast<unsigned long long>(std::pow(16, D));
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const unsigned long long SHIFT = 4 * D;
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const unsigned long long MASK = M - 1;
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unsigned long long s = 0;
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unsigned long long k = 0;
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while (k <= n)
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{
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unsigned long long r = 8 * k + j;
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unsigned long long v = static_cast<unsigned long long>(std::pow(16ul, n - k)) % r;
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s = (s + (v << SHIFT) / r) & MASK;
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k += 1;
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}
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unsigned long long t = 0;
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k = n + 1;
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while (true)
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{
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unsigned long long xp = static_cast<unsigned long long>(powneg(16, n - k) * M);
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unsigned long long newt = t + xp / (8 * k + j);
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if (t == newt)
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break;
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else
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t = newt;
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k += 1;
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}
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return s + t;
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}
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unsigned long long piDigitHex(unsigned long long n) {
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const unsigned long long D = 14;
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const unsigned long long M = static_cast<unsigned long long>(std::pow(16, D));
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const unsigned long long SHIFT = 4 * D;
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const unsigned long long MASK = M - 1;
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n -= 1;
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unsigned long long x = (4 * s(1, n) - 2 * s(4, n) - s(5, n) - s(6, n)) & MASK;
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return x;
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}
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