Program for Analyzing alien PI computing

There is a program that uses only four lines of code to calculate pi, known as the alien PI computing program.

Many people discuss and analyze the implementation principle of the program, such as: http://blog.csdn.net/panqiaomu/archive/2006/05/07/711776.aspx

But I always felt that the analysis was not thorough enough, so I analyzed it myself.

1. modify the original program to a more understandable form;

2. Use the same algorithm to calculate pi in an Excel file.

Download (note the modified extension)
 

/* File: pi800.c Name: Program for Analyzing alien PI computing Author: zyl910 Blog: http://blog.csdn.net/zyl910/ Version: V1.0 Updata: 2006-11-5 Program for Analyzing alien PI computing ~~~~~~~~~~~~~~~~~~~~~~ Original program: Int A = 10000, B, c = 2800, d, e, f [2801], G; Main (){ For (; B-c ;) F [B ++] = A/5; For (; D = 0, G = C * 2; C-= 14, printf ("%. 4D", e + D/A), E = D %) For (B = C; D + = f [B] * A, F [B] = D % -- g, D/= g --, -- B; D * = B ); } Pi formula: 1 2 3 K Pi = 2 + --- * (2 + --- * (2 + --- * (2 +... (2 + ---- * (2 + ...))...))) 3 5 7 2 k + 1 */ Int main (void) { Long A = 10000; // Scaling Factor Long c = 2800; // number of iterations Long f [2801]; // intermediate Calculation Result Long D; // cumulative inner cycle error Long E = 0; // cumulative error of external circulation Long B, G; // molecules and denominator. K/(2 k + 1) Int I, J; For (I = 0; I <C; I ++) // If you fully follow the formula, the cycle condition should be "for (I = 1; I <= C; I ++) ". Fortunately, F [2800] has a very small contribution, and getting any value will not have a big impact on accuracy. F [I] = 2 * A/10; // 2 is the coefficient 2 in the formula. A/10 indicates that a decimal bit is retained, because the output starts from 3. While (C> 0) { D = 0; G = (C * 2 + 1)-2; // denominator. Because 4 bits are output in each loop, we multiply the number of bits in the subsequent operation by a, so here we have to-2 B = C; // molecule While (B> 0) { /* Multiply by the numerator according to the formula */ D * = B; D + = f [B] * A; // because each External Loop outputs four digits /* Divide by the denominator according to the formula */ F [B] = D % G; // part of the numerator with a score D/= g; // an integer with a score /* Next */ G-= 2; B --; } Printf ("%. 4D", e + D/); E = D %; C-= 14; // because the precision is fixed to 800 bits, after every 4 bits are output, it is equivalent to reducing the precision requirement by 4 bits, so 14 items can be counted at a time } Printf ("/N "); Return 0; }