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
难点是，数据类型最长有32位（4字节或者2字），数值范围是-2147483648~2147483648或者0~4294967295，但题目中指出输入数据位数长度可以达到1000位，10^999>>4294967295,故不能用常规方法
具体解决方法是，将数字利用字符串的形式表示，每个字符都是数字，1000个连续字符也没问题，再将两个不同字符串相加得到最终结果。
有一次提交时，出现了“Presentation
Error”的错误，缘由是输出结果的格式不符合要求，比方少个空格什么的。
1 #include <iostream>
2 #include <string>
3 using namespace std;
4 int main()
5 {
6 int n;
7 while(cin>>n)//n为case数
8 {
9 for(int i=1;i<=n;i++)
10 {
11 string a,b,c;//3个字符串
12 cin>>a>>b;
13 int la=a.length()-1,lb=b.length()-1,jw=0,ta,tb,tt,f=0;
14 char tc;
15 while(la>=0||lb>=0)
16 {
17
18 if(la<0) ta=0;
19 else ta=a[la]-‘0‘;
20 if(lb<0) tb=0;
21 else tb=b[lb]-‘0‘;
22 tt=jw+ta+tb;//tt为a和b两位相加结果
23 jw=tt/10;//jw为进位
24 tc=tt%10+‘0‘;//tc为赋值给字符串c之前的一个中转
25 if(tc!=‘0‘) f=1;//f为进位标志
26 c+=tc;
27 la--;lb--;
28 }
29 if(jw>0)
30 {
31 f=1;
32 tc=jw+‘0‘;
33 c+=tc;
34 }
35
36 if(i!=1) cout<<endl;
37 cout<<"Case "<<i<<":"<<endl;
38 cout<<a<<" + "<<b<<" = ";
39 if(f==1)
40 {
41 for(int j=c.length()-1;j>=0;j--)
42 cout<<c[j];
43 cout<<endl;
44 }
45 else cout<<0<<endl;
46 }
47 }
48 return 0;
49 }