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
思路：这题除了格式有点坑以外还行，我这里把大数加法用个函数表示，显得更清晰明了......
1 #include<iostream>
2 #include<cstring>
3 #include<cstdio>
4 #include<cmath>
5 #include<algorithm>
6 #include<map>
7 #include<vector>
8 #include<set>
9 #include<queue>
10 using namespace std;
11 #define ll long long
12
13 string Sum(string a,string b)//大数加法
14 {
15 //补前导零
16 while(a.size()<b.size())
17 a.insert(0,"0");
18 while(b.size()<a.size())
19 b.insert(0,"0");
20
21 string ans="";//记录结果
22
23 int jinwei=0,sum,yu;//运算所需
24
25 for(int i=a.size()-1;i>=0;i--)//从末尾算起
26 {
27 sum=(a[i]-'0')+(b[i]-'0')+jinwei;
28 jinwei=sum/10;
29 yu=sum%10;
30 ans+=(yu+'0');//加上这一位的余数
31 }
32 if(jinwei)//可能多一位
33 ans+=(jinwei+'0');
34
35 reverse(ans.begin(),ans.end());//由于是从末尾算起，需要逆置字符串
36
37 return ans;
38 }
39
40 int main()
41 {
42 ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
43
44 int T;
45
46 cin>>T;
47
48 string a,b,s;
49
50 for(int k=1;k<=T;k++)
51 {
52 cin>>a>>b;
53 s=Sum(a,b);
54
55 cout<<"Case "<<k<<":"<<endl;
56 cout<<a<<" + "<<b<<" = "<<s<<endl;
57
58 if(k!=T)
59 cout<<endl;
60 }
61 return 0;
62 }