HDU1002 -A + B Problem II(大数a+b)

　　

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 315214    Accepted Submission(s): 61139

Problem Description

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

 

Author

Ignatius.L

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int main() {
    int t,i,k;
    cin>>t;
    for(k=1; k<=t; k++) {
        string a,b,c,m,n;
        int sum,add;
        cin>>a>>b;
        m=a;
        n=b;
        int lena=a.length();
        int lenb=b.length();
        reverse(a.begin(),a.end());//反转字符串 
        reverse(b.begin(),b.end());
        add=0;
        //模拟加法运算 
        for(i=0; i<lena||i<lenb; i++) {
            if(i<lena&&i<lenb)
                sum=a[i]-'0'+b[i]-'0'+add;
            else if(i<lena)
                sum=a[i]-'0'+add;
            else if(i<lenb)
                sum=b[i]-'0'+add;
            add=0;
            if(sum>9) {
                add=1;
                sum-=10;
            }
            c+=sum+'0';
        }
        if(add)
            c+=add+'0';
        reverse(c.begin(),c.end());//还原 
        cout<<"Case "<<k<<":"<<endl;
        cout<<m<<" + "<<n<<" = ";
        cout<<c<<endl;
        if(k!=t)
            cout<<endl;
    }
    return 0;
}