为π(pi)实现Spigot算法

本文介绍了为π(pi)实现Spigot算法的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我很难理解此处位于页面底部.

I'm having a hard time understanding the Spigot algorithm for π (pi) found here at the bottom of the page.

在第2部分将A放入常规格式"的底部，我迷路了，我不确定如何在 C (或任何语言)中实现它

I'm getting lost at the bottom of part 2 "Put A into regular form", I'm not exactly sure how to implement this in C (or any language really)

推荐答案

#include <math.h> #include <stdio.h> #define N 100 int len = floor(10 * N/3) + 1; int A[len]; for(int i = 0; i < len; ++i) { A[i] = 2; } int nines = 0; int predigit = 0; for(int j = 1; j < N + 1; ++j) { int q = 0; for(int i = len; i > 0; --i) { int x = 10 * A[i-1] + q*i; A[i-1] = x % (2*i - 1); q = x / (2*i - 1); } A[0] = q%10; q = q/10; if (9 == q) { ++nines; } else if (10 == q) { printf("%d", predigit + 1); for (int k = 0; k < nines; ++k) { printf("%d", 0); } predigit, nines = 0; } else { printf("%d", predigit); predigit = q; if (0 != nines) { for (int k = 0; k < nines; ++k) { printf("%d", 9); } nines = 0; } } } printf("%d", predigit);