<pre name="code" class="cpp">#include "oj.h"
#include<iostream>

int main(void)
{
string strMultiplierA = "9999";
string strMultiplierB = "9999";
string strRst = "\0";
multiply (strMultiplierA,strMultiplierB,strRst);
cout << strRst;
return 0;
}

/*****************************************************************************
Prototype    : multiply
Description  : 两个任意长度的长整数相乘, 输出结果
Input pAram  : 
const std::string strMultiplierA  乘数A
const std::string strMultiplierB  乘数B
Output       : 
std::string strRst            乘法结果
Return Value : 
int                       0  正确  
-1  异常
*****************************************************************************/
int multiply (const std::string strMultiplierA,const std::string strMultiplierB, std::string &strRst) 
{

	/* 在这里实现功能 */
	string chengshuA = strMultiplierA;
	string chengshuB = strMultiplierB;
	int a = chengshuA.length();
	int b = chengshuB.length();
	int strRst_length = 0;
	int c = (a+1)*(b+1);
	int *p = new int[c];//逆序存放的结果
	int *pA = new int[a];//逆序存放的乘数A
	int *pB = new int[b];//逆序存放的乘数B

	//测试是否有乘数为空
	if ((a == 0) || (b == 0))
		return -1;

	for (int i = 0; i != c; i++)
		p[i] = 0;

	//把乘数逆序放到数组pA中，若chengshuA是12345，则pA是54321
	for (string::size_type index = 0; index !=  chengshuA.length(); index++) 
		pA[a-1-index] = chengshuA.at(index) - '0';//pA[(a-1) - index]


	for (string::size_type index = 0; index != chengshuB.length(); index++)
		pB[b-1-index] = chengshuB.at(index) - '0';


	for (int temp_b = 0; temp_b != b; temp_b++)  //每个位循环相乘
	{
		for (int temp_a = 0; temp_a != a; temp_a++)//pA的每个位循环与pB[temp_b]相乘
		{
			//p[temp_a+temp_b]利用temp_b起到移位的作用，如temp_b是十位数，则在乘法计算中要向后移动移位。
			int temp = p[temp_a+temp_b] + pA[temp_a]*pB[temp_b];
			p[temp_a+temp_b] = temp % 10;
			int carry = temp/10; //进位

			int x = temp_a + temp_b +1;
			//c = (a+1)*(b+1),足够大;大数相加与数相乘同时计算,进位有可能是多位数
			while(carry != 0)//进位不等于0
			{
				p[x] = p[x] + carry%10;//若前面有进位，则提前把进位加到求和结果p[x]上
				carry = carry/10;
				x++;
			}
		}
	}

	while (c-- > 0)  //判断结果有几位
	{
		if (p[c] != 0)
		{
			strRst_length = c + 1;
			break;
		}
	}

	char ch;
	for (int i = strRst_length - 1; i >= 0 ; i--)  //把结果放入字符串中
	{
		ch = p[i] + '0';
		strRst.push_back(ch);
	}
	if (strRst.empty())//如果结果为0，则输出字符串为“0”
	{
		strRst = "0";
	}

	return 0;
}