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

---

高精度加法模板，注意前导0不输出。

---

// 高精度加法 
#include<iostream>
#include<cstring>
using namespace std;
int a[1005],b[1005];

int pluss(int a[],int b[]){
    int l;
    l=a[0]>b[0]?a[0]:b[0];
    for(int i=1;i<=l;i++){
        a[i+1]+=(a[i]+b[i])/10;
        a[i]=(a[i]+b[i])%10;
    }
    if(a[l+1]>0) a[0]=l+1;
    else a[0]=l;
    return 0;
}

int main(){
    int T;
    scanf("%d",&T);
    for(int cnt=1;cnt<=T;cnt++){
        
        //读入 
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        string s1,s2;
        cin>>s1>>s2;
        a[0]=s1.length();
        b[0]=s2.length();
        for(int i=1;i<=a[0];i++){
            a[i]=s1[a[0]-i]-'0';//倒序存储 
        }
        for(int i=1;i<=b[0];i++){
            b[i]=s2[b[0]-i]-'0';
        } 
        
        cout<<"Case "<<cnt<<":"<<endl<<s1<<" + "<<s2<<" = ";
        
        pluss(a,b);
        
        int flag=0;
        for(int i=a[0];i>=1;i--){
            if(flag==0&&a[i]==0) continue;//坑点：前导0不输出。 
            else{
                cout<<a[i];
                flag=1;
            }
        }

        cout<<endl;
        if(cnt!=T) cout<<endl;
        
    }
    return 0;
}