hdu 1002 大数相加

I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.
Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.
Output
For each test case, you should output two lines. The first line is “Case #:”, # means the number of the test case. The second line is the an equation “A + B = Sum”, Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.
Sample Input
2
1 2
112233445566778899 998877665544332211
Sample Output
Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

高精度加法模板题
不过由于自己太菜了，写了好久加改了好久才过的
AC代码：

#include<iostream>
#include<algorithm>
#include<string>
#include<cstring>
using namespace std;
int main(){
	int n;
	string a,b;
	int aa[2005],bb[2005],p[2005];
	cin>>n;
	for(int t=1;t<=n;t++){
		cin>>a>>b;
		int aaa=a.size();
		int bbb=b.size();
		memset(aa,0,sizeof(aa));
		memset(bb,0,sizeof(bb));
		memset(p,0,sizeof(p));
		cout<<"Case "<<t<<":\n"<<a<<" + "<<b<<" = ";
		int i,k,ans=0;
		for(i=aaa-1,k=0;i>=0;k++,i--){
			aa[k]=a[i]-'0';
		}
		for(i=bbb-1,k=0;i>=0;k++,i--){
			bb[k]=b[i]-'0';
		}
		int len=max(aaa,bbb),jingwei=0;
		for(int j=0;j<len;j++){
			p[j]=aa[j]+bb[j]+jingwei;
			jingwei=0;
			if(p[len-1]>=10)len++;
			if(p[j]>=10){
				p[j]%=10;
				jingwei=1;
			}
		}
		for(int j=len-1;j>=0;j--)
		cout<<p[j];
		cout<<endl;	
		if(t<n)cout<<endl;			
	}return 0;
}