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

1 2

112233445566778899 998877665544332211

Sample Output

Case 1: 1 + 2 = 3

Case 2: 112233445566778899 + 998877665544332211 = 1111111111111111110

Author

Ignatius.L

 1 #include<iostream>
 2 #include<string>
 3 #include<algorithm>
 4 using namespace std;
 5 int main() {
 6     int t,i,k;
 7     cin>>t;
 8     for(k=1; k<=t; k++) {
 9         string a,b,c,m,n;
10         int sum,add;
11         cin>>a>>b;
12         m=a;
13         n=b;
14         int lena=a.length();
15         int lenb=b.length();
16         reverse(a.begin(),a.end());//反转字符串 
17         reverse(b.begin(),b.end());
18         add=0;
19         //模拟加法运算 
20         for(i=0; i<lena||i<lenb; i++) {
21             if(i<lena&&i<lenb)
22                 sum=a[i]-'0'+b[i]-'0'+add;
23             else if(i<lena)
24                 sum=a[i]-'0'+add;
25             else if(i<lenb)
26                 sum=b[i]-'0'+add;
27             add=0;
28             if(sum>9) {
29                 add=1;
30                 sum-=10;
31             }
32             c+=sum+'0';
33         }
34         if(add)
35             c+=add+'0';
36         reverse(c.begin(),c.end());//还原 
37         cout<<"Case "<<k<<":"<<endl;
38         cout<<m<<" + "<<n<<" = ";
39         cout<<c<<endl;
40         if(k!=t)
41             cout<<endl;
42     }
43     return 0;
44 }

HDU1002 -A + B Problem II(大数a+b) A + B Problem II

相关推荐