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=; 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=;
         //模拟加法运算
         for(i=; i<lena||i<lenb; i++) {
             if(i<lena&&i<lenb)
                 sum=a[i]-''+b[i]-''+add;
             else if(i<lena)
                 sum=a[i]-''+add;
             else if(i<lenb)
                 sum=b[i]-''+add;
             add=;
             if(sum>) {
                 add=;
                 sum-=;
             }
             c+=sum+'';
         }
         if(add)
             c+=add+'';
         reverse(c.begin(),c.end());//还原
         cout<<"Case "<<k<<":"<<endl;
         cout<<m<<" + "<<n<<" = ";
         cout<<c<<endl;
         if(k!=t)
             cout<<endl;
     }
     return ;
 }