java - 从python到java的并行赋值变量(Pi算法)

Question

我想将 python 算法转换为 Java，我有这个源代码(使用并行赋值变量(在 Java 中不存在 :( )

# -*- coding: cp1252 -*-
#! /usr/bin/env python

import sys



def main():
    k, a, b, a1, b1 = 2L, 4L, 1L, 12L, 4L
    while 1:
        p, q, k = k*k, 2L*k+1L, k+1L
        a, b, a1, b1 = a1, b1, p*a+q*a1, p*b+q*b1
        d, d1 = a/b, a1/b1
        while d == d1:
            output(d)
            a, a1 = 10L*(a%b), 10L*(a1%b1)
            d, d1 = a/b, a1/b1

def output(d):
    sys.stdout.write(`int(d)`)
    sys.stdout.flush()
    #ecriture en continue du chiffre
    pi = open("flypi.html", "a")
    pi.write(`int(d)`)
    pi.write("\n")
    pi.close()


main()

所以，首先我重新编写了没有并行赋值变量的相同脚本:

# -*- coding: cp1252 -*-
#! /usr/bin/env python

import sys



def main():
    #k, a, b, a1, b1 = 2L, 4L, 1L, 12L, 4L
    k = 2L
    a = 4L
    b = 1L
    a1 = 12L
    b1 = 4L

    while 1:

        #p, q, k = k*k, 2L*k+1L, k+1L
        kk = k
        p = kk*kk
        q = 2L*kk+1L
        k = kk+1L

        #a, b, a1, b1 = a1, b1, p*a+q*a1, p*b+q*b1
        aa = a
        bb = b
        a = a1
        b = b1
        a1 = p*aa+q*a1
        b1 = p*bb+q*b1

        #d, d1 = a/b, a1/b1
        d = a/b
        d1 = a1/b1
        while d == d1:
            output(d)

            #a, a1 = 10L*(a%b), 10L*(a1%b1)
            a = 10L*(a%b)
            a1 = 10L*(a1%b1)

            #d, d1 = a/b, a1/b1
            d = a/b
            d1 = a1/b1

def output(d):
    sys.stdout.write(`int(d)`)
    sys.stdout.flush()
    #ecriture en continue du chiffre
    pi = open("flypi.html", "a")
    pi.write(`int(d)`)
    pi.write("\n")
    pi.close()


main()

这两个脚本的输出是一样的:

31415926535897932384626433832795028841971693993751058209749445923078164062862089 (crt+c)

现在这是我用 Java 编写的脚本(几乎与第二个 python 脚本相同):

public static void cal(){
    //int i = 0;
    long d = 0;
    long k = 2L;
    long a = 4L;
    long b = 1L, a1 = 12L, b1 = 4L;
    long p = 0, q = 0, d1 = 0;

    long aa = 0, bb = 0;
    long kk = 0;


    while(true){

        kk = k;
        p = kk*kk;
        q = 2L*kk+1L;
        k = kk+1L;

        aa = a;
        bb = b;
        a = a1;
        b = b1;
        a1 = p*aa+q*a1;
        b1 = p*bb+q*b1;

        d = a/b;
        d1 = a1/b1;
        while(d == d1){
            System.out.print(d);
            try {
                Thread.sleep(100);
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
            a = 10L*(a%b);
            a1 = 10L*(a1%b1);

            d = a/b;
            d1 = a1/b1;
        }
        //i++;

    }
}

但是输出是错误的:

31415926530000000001-100000000000000000100-300000101000000000000000000000000000000000000 (ctr+c)

谢谢，很抱歉发了这么长的帖子:)

编辑:所以是的，这是缓冲区溢出。我尝试实现 BigInteger 并且效果很好!谢谢!

Answer 1

在 Python 中整数可以是任意大的。在 Java 中，long 由 64 位组成，因此只能存储小于 2**64/2 的数字。

如果一个数字太大，它的前几位将被丢弃，最高有效位不是覆盖整数的符号，导致在数学上不可能的地方出现负数。

按照 ajb 的建议使用 BigInteger 或以某种方式更改您的计算。