大数乘法：计算两个任意长度数字的积

//大数乘法算法实现
#include<iostream>
#include<string>
#include<cstring>
using?namespace?std;

int?main()
{
????string?num1,num2;
????cin?>>?num1?>>?num2;
????cout?<<?num1.size()?<<?"?"?<<?num2.size()?<<?endl;
????const?char*?n1;
????const?char*?n2;
????if?(num1.size()?<?num2.size())
????{
????????n1?=?num2.c_str();
????????n2?=?num1.c_str();
????}
????else
????{
????????n1?=?num1.c_str();
????????n2?=?num2.c_str();
????}
????char*?n?=?new?char[strlen(n1)+strlen(n2)+1];
????for?(unsigned?int?i?=?0;?i?<?strlen(n1)+strlen(n2);?i++)
????????n[i]?=?'0';
????n[strlen(n1)+strlen(n2)]='';
????cout?<<?strlen(n)?<<?endl;
????
????int?count?=?0,flag?=?0;
????for?(int?i?=?strlen(n1)-1;?i?>=?0;?i--)
????{
????????flag++;
????????int?x1?=?n1[i]-'0';
???????//?cout?<<?"n1["<<?i?<<?"]为："?<<?x1?<<?endl;
????????char?carry?=?'0';
????????for?(int?j?=?strlen(n2)-1;?j?>=?0;?j--)
????????{
????????????int?x2?=?n2[j]-'0';
???????????//?cout?<<?"n2["<<?j?<<?"]为："?<<?x2?<<?endl;
????????????//cout?<<?"当前位未改变前值为:?"?<<?n[count]?<<?endl;
????????????int?sum?=?x1*x2?+?(carry-'0')?+?n[count]-'0';
????????????//cout?<<?"sum?is?"?<<?sum?<<?endl;
????????????n[count++]?=?(sum?%?10)+'0';
????????????carry?=?(sum?/?10)+'0';
???????????//?cout?<<?"当前位的值为:?"?<<?n[count-1]?<<?endl;
???????????//?cout?<<?"carry的值为："?<<?carry?<<?endl;
????????}
????????if?(carry?!=?'0')
????????{
????????????n[count]?=?carry;
????????????count?=?flag;
????????????//cout?<<?"当前位的值为:?"?<<?n[count]?<<?endl;
????????}
????????else
????????????count?=?flag;
????}

????for?(int?i?=?strlen(n)-1;?i?>=?0;?i--)
????{
????????if?((i?==?strlen(n)-1)&&(n[i]?==?'0'))
????????????continue;
????????cout?<<?n[i];
????}
????cout?<<?endl;
????delete[]n;
????system("pause");
????return?0;
}