实现一个计算大位数（如100位以上）相乘结果的函数（请完全用算法实现）
#include<iostream.h>
#include<string.h>
#include<stdio.h>
multiply(char * a_strMultp1,char * a_strMultp2)
{
int Len1; // the length of multiplier1
int Len2; // the length of multiplier2
Len1 = strlen(a_strMultp1);
Len2 = strlen(a_strMultp2);
char *strRet= new char(iMultp1Len+iMultp2Len); // the result;
memset(strRet,'0',iMultp1Len+iMultp2Len);
// if either's length is 0,then exit;
if(Len1 <= 0 || Len 2<= 0)
{
cout<<"error"<<endl;
exit(0);
}
int i;
int j;
int iCarry; // the Carry;
int iDigit; // the Digit;
iCarry = 0;
iDigit = 0;
int len;
int temp = 0;
for(i = Len2-1 ; i >=0 ; i --)
{
for(j = Len1-1; j >=0; j--)
{ int jj =Len2+Len1-j-i-2;
iDigit = (a_strMultp2[i] - '0')*(a_strMultp1[j] - '0') + iCarry;
iCarry = iDigit / 10;
iDigit = iDigit % 10;
temp = strRet[jj] - '0' +iDigit;
if (temp/10)
iCarry = iCarry + temp/10;
strRet[jj]= temp%10 + '0';
if (iCarry)
strRet[jj+1]=iCarry%10 + strRet[jj+1];
iCarry = iCarry /10;
}
}
if (iCarry)
{
strRet[Len2+Len1-1]= iCarry + '0';
len = Len2+Len1-1;
}
else
len = Len2+Len1-2;
for (i = len;i>=0;i--)
cout<<strRet[i];
cout<<endl;
}
main()
{
multiply("177856","196196");
}
