[讨论] 各种语言计算圆周率的程序

dcexpert 2020-4-2 10:18 楼主

无意中看到一个介绍各种语言计算任意精度圆周率的网站，常见的编程语言都有，分享给大家。

10 PRINT CHR$(147)
20 n = 100
30 ln = int(10*n/4)
40 nd = 1
50 dim a(ln)
60 n9 = 0
70 pd = 0 :rem First predigit is a 0
80 :
90 for j = 1 to ln
100    a(j-1) = 2 :rem Start with 2s
110 next j
120 :
130 for j = 1 to n
140     q = 0
150     for i = ln to 1 step -1 :rem Work backwards
160         x = 10*a(i-1) + q*i
170         a(i-1) = x - (2*i-1)*int(x/(2*i-1)) :rem X - INT ( X / Y) * Y
180         q = int(x/(2*i - 1))
190     next i
200     a(0) = q-10*int(q/10)
210     q = int(q/10)
220     if q=9 then n9 = n9 + 1 : goto 450
240     if q&lt;&gt;10 then 350
250     rem q == 10
260        d = pd+1 : gosub 500
270        if n9 &lt; 0 then 320
280           for k = 1 to n9
290              d = 0: gosub 500
300           next k
310        rem end if
320        pd = 0
330        n9 = 0
335        goto 450
340     rem q &lt;&gt; 10
350        d = pd: gosub 500
360        pd = q
370        if n9 = 0 then 450
380           for k = 1 to n9
390              d = 9 : gosub 500
400           next k
410           n9 = 0
450 next j
460 print mid$(str$(pd),2,1)
470 end
480 :
490 rem outputd
500 if nd=0 then print mid$(str$(d),2,1); : return
510 if d=0 then return
520 print mid$(str$(d),2,1);".";
530 nd = 0
550 return

def calcPi():
    q, r, t, k, n, l = 1, 0, 1, 1, 3, 3
    while True:
        if 4*q+r-t &lt; n*t:
            yield n
            nr = 10*(r-n*t)
            n  = ((10*(3*q+r))//t)-10*n
            q  *= 10
            r  = nr
        else:
            nr = (2*q+r)*l
            nn = (q*(7*k)+2+(r*l))//(t*l)
            q  *= k
            t  *= l
            l  += 2
            k += 1
            n  = nn
            r  = nr
 
import sys
pi_digits = calcPi()
i = 0
for d in pi_digits:
    sys.stdout.write(str(d))
    i += 1
    if i == 40: print(""); i = 0

type AnyWriteableObject={write:((textToOutput:string)=>any)};
 
function calcPi(pipe:AnyWriteableObject) {
    let q = 1n, r=0n, t=1n, k=1n, n=3n, l=3n;
    while (true) {
        if (q * 4n + r - t < n* t) {
            pipe.write(n.toString());
            let nr = (r - n * t) * 10n;
            n  = (q * 3n + r) * 10n / t - n * 10n ;
            q  = q * 10n;
            r  = nr;
        } else {
            let nr = (q * 2n + r) * l;
            let nn = (q * k * 7n + 2n + r * l) / (t * l);
            q = q * k;
            t = t * l;
            l = l + 2n;
            k = k + 1n;
            n  = nn;
            r  = nr;
        }
    }
}
 
calcPi(process.stdout);