Haut OJ 1357: lunch question (I) -- high precision multiplication

Problem description ：

   12 It's o'clock again , It's time for lunch , Small C  Collapse in bed and don't want to get up , Small A  Don't want to get used to him , So I said to him ：“ Small C, You didn't make it last week choice  Sister's addition , Now you can make choice  Sister's multiplication , I'll bring you some rice .” Can you help me C.

Input ：

   Two integers  A  and  B,A and B The length of l（0 <= l <= 1000).

Output ：

Multi instance testing , Output  A*B  Result .

The sample input ：

4038
5255
55718
7221
8160
104906
3511906
8767574

Sample output ：

21219690
402339678
856032960
30790895736044

---

Cause analysis ：

1. Create three int Array , Initialize to 0, Record length , Save each number in the string into the array cell （ Input backwards ）

2. Multiply each number ,     c[i+j-1] + =a[i]*b[j];        c[i+j]+=c[i+j-1]/10;  Don't let it slip +

3. Remove leading 0
              

---

Solution ：

#include<bits/stdc++.h>
using namespace std;
#define endl "\n"
long long a[1000005];
int main(void)
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    string str1,str2;
    while(cin>>str1>>str2)
 {

    int a[2000],b[2000],c[5000],i,j,len;
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    memset(c,0,sizeof(c));
    a[0]=str1.length();
    b[0]=str2.length();
    for(i=1;i<=a[0];i++)
        a[i]=str1[a[0]-i]-'0';
    for(i=1;i<=b[0];i++)
        b[i]=str2[b[0]-i]-'0';
    for(i=1;i<=a[0];i++)
        for(j=1;j<=b[0];j++)
    {
        c[i+j-1]+=a[i]*b[j];
        c[i+j]+=c[i+j-1]/10;
        c[i+j-1]%=10;

    }
    len=a[0]+b[0]+1;
    while(len>1&&c[len]==0)
        len--;

    for(i=len;i>=1;i--)
        cout<<c[i];
        cout<<endl;


 }

    return 0;
}