HDU-1002-A + B Problem II(高精度加法)

I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B. 

InputThe 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. 
OutputFor 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

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高精度加法模板，注意前导0不输出。

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 1 // 高精度加法 
 2 #include<iostream>
 3 #include<cstring>
 4 using namespace std;
 5 int a[1005],b[1005];
 6 
 7 int pluss(int a[],int b[]){
 8     int l;
 9     l=a[0]>b[0]?a[0]:b[0];
10     for(int i=1;i<=l;i++){
11         a[i+1]+=(a[i]+b[i])/10;
12         a[i]=(a[i]+b[i])%10;
13     }
14     if(a[l+1]>0) a[0]=l+1;
15     else a[0]=l;
16     return 0;
17 }
18 
19 int main(){
20     int T;
21     scanf("%d",&T);
22     for(int cnt=1;cnt<=T;cnt++){
23         
24         //读入 
25         memset(a,0,sizeof(a));
26         memset(b,0,sizeof(b));
27         string s1,s2;
28         cin>>s1>>s2;
29         a[0]=s1.length();
30         b[0]=s2.length();
31         for(int i=1;i<=a[0];i++){
32             a[i]=s1[a[0]-i]-'0';//倒序存储 
33         }
34         for(int i=1;i<=b[0];i++){
35             b[i]=s2[b[0]-i]-'0';
36         } 
37         
38         cout<<"Case "<<cnt<<":"<<endl<<s1<<" + "<<s2<<" = ";
39         
40         pluss(a,b);
41         
42         int flag=0;
43         for(int i=a[0];i>=1;i--){
44             if(flag==0&&a[i]==0) continue;//坑点：前导0不输出。 
45             else{
46                 cout<<a[i];
47                 flag=1;
48             }
49         }
50 
51         cout<<endl;
52         if(cnt!=T) cout<<endl;
53         
54     }
55     return 0;
56 }