众所周知，计算机中int用32位表示，可表示范围0 - 2^32，约42亿， long用64位表示，可表示范围为0-2^64 约184亿亿。如果运算是超出这个范围，就会溢出，即无法获取正确的计算结果。这时候就需要一种特殊的计算程序。本文通过Python语言，实现了一个简单的大数计算程序：通过输入两个任意长度的由数字组成的字符串，进行+ - * /计算，计算结果也是一个字符串。

1.代码

2.代码

代码如下：

bigdata_calc.py

#! /usr/bin/python
# -*- coding: utf-8 -*-
import sys

def get_10(c):
	return ord(c)-ord(‘0‘)

def add_one_bit(n1,n2,n3=0):
	a = get_10(n1)
	b = get_10(n2)
	
	sum = a+b+n3
	
	yu = sum%10
	shang = (sum-yu)/10
	
	return str(yu),shang

def add(n1,n2):
	
	l1=len(n1)
	l2=len(n2)
	
	s1 = n1
	s2 = n2
	if l1>l2:
		s1 = n2
		s2 = n1
	
	l1 = len(s1)
	l2 = len(s2)
	
	sum = []	
	shang = 0
	for i in range(l1):
		c,shang = add_one_bit(s1[l1-1-i],s2[l2-1-i],shang)
		sum.append(c)

	for i in range(l1,l2):
		c,shang = add_one_bit(‘0‘,s2[l2-1-i],shang)
		sum.append(c)
	if shang > 0:
		sum.append(str(shang))
		
	for i in range(len(sum)/2):
		tmp = sum[i]
		sum[i]=sum[len(sum)-1-i]
		sum[len(sum)-1-i] = tmp
	return sum	

def mul(n1,n2):
	l = len(n1)
	sum=[]
	for i in range(l):
		v = get_10(n1[i])
		if v>0:
			sum_tmp = n2
			for j in range(l-1-i):
				sum_tmp += "0"
			add_v = sum_tmp
			for j in range(1,v):
				sum_tmp = add(sum_tmp,add_v)
			if i==0:
				sum = sum_tmp
			else:
				sum = add(sum,sum_tmp)
	return sum

def sub_one_bit(n1,n2,n3):
	a = get_10(n1)
	b = get_10(n2)
	
	sub = a-b-n3
	
	jie=0
	while sub<0:
		jie += 1
		sub += 10
	return str(sub),jie

def large(n1,n2):
	if len(n1)>len(n2):
		return True
	elif len(n1)<len(n2):
		return False
	else:
		for i in range(len(n1)):
			if ord(n1[i])>=ord(n2[i]):
				return True
		return False
		
def sub(n1,n2):
	
	s1 = n1
	s2 = n2
	flag=False
	if not large(n1,n2):
		s1 = n2 
		s2 = n1
		flag = True

	l1 = len(s1)
	l2 = len(s2)
	
	sub = []	
	jie = 0
	for i in range(l2):

			c,jie = sub_one_bit(s1[l1-1-i],s2[l2-1-i],jie)
			sub.append(c)
			
	has_jie = 0		
	if jie>0:
		c,jie = sub_one_bit(n1[l2-1-l1],str(jie),0)
		sub.append(c)
		has_jie=1
	
	while has_jie<l1-l2:
		sub.append(s1[l1-l2-1-has_jie])
		has_jie += 1
		
	if flag:
		sub.append(‘-‘)
		
	for i in range(len(sub)/2):
		tmp = sub[i]
		sub[i]=sub[len(sub)-1-i]
		sub[len(sub)-1-i] = tmp
	return sub	

def div(n1,n2):
	return "Todo..."
	
def calc(n1,op,n2):
	if op == ‘+‘:
		return add(n1,n2)
	elif op == ‘-‘:
		return sub(n1,n2)
	elif op == ‘*‘:
		return mul(n1,n2)
	elif op == ‘/‘:
		return div(n1,n2)
	else:
		return "unsupported operation"
	
if __name__ == ‘__main__‘:
	if len(sys.argv)<4:
		print "Usage: bigdata_calc.py  num1 {+,-,*,/} num2"
	else:
		res = calc(sys.argv[1],sys.argv[2],sys.argv[3])
		
		print "%50s\n%s\n%50s\n=\n%50s" %(sys.argv[1],sys.argv[2],sys.argv[3],"".join(res))
		
		
		

            虽然个人进行了测试，但难免还有bug，请指正。

3.测试

测试一：

123

*

456

=

56088

测试二：

123456789123456789123456789123456789123456789123456789123456789123456789123456789123456789

*

123456789123456789123456789123456789123456789123456789123456789123456789123456789123456789

=

15241578780673678546105778311537878076969977842402077607834177373266277138698376
90413047663907941887364731910821521934278311957735101981191892004648682028105472
0515622620750190521

结果如下：

Python编程：以String方式进行大数计算,布布扣,bubuko.com

Python编程：以String方式进行大数计算