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
解题思路：大数加法，简单模拟竖式运算或者直接调用java中BigInteger大数类的add()方法即可。
AC之C++代码：
1 #include<bits/stdc++.h>
2 using namespace std;
3 int main(){
4 string a,b,r;int t,tmp;bool flag;
5 while(cin>>t){
6 flag=true;
7 for(int i=1;i<=t;++i){
8 cin>>a>>b;tmp=0;r="";
9 if(!flag)cout<<endl;
10 cout<<"Case "<<i<<":\n"<<a<<" + "<<b<<" = ";
11 if(a.size()>b.size())swap(a,b);
12 for(int i=a.size()-1,j=b.size()-1;~j;--i,--j){
13 int now=(i>=0?a[i]-‘0‘:0)+b[j]-‘0‘+tmp;
14 tmp=now>9?1:0;now%=10;
15 r+=‘0‘+now;
16 }
17 if(tmp)r+=‘1‘;
18 for(int i=r.size()-1;~i;--i)cout<<r[i];
19 cout<<endl;flag=false;
20 }
21 }
22 return 0;
23 }
AC之java代码：
1 import java.util.Scanner;
2 import java.math.BigInteger;
3 public class Main {
4 public static void main(String[] args) {
5 Scanner scan = new Scanner(System.in);
6 int t=scan.nextInt();
7 for(int i=1;i<=t;++i){
8 BigInteger a = scan.nextBigInteger();
9 BigInteger b = scan.nextBigInteger();
10 System.out.println("Case "+i+":");
11 System.out.println(a+" + "+b+" = "+a.add(b));
12 if(i!=t)System.out.println();
13 }
14 }
15 }
原文地址：https://www.cnblogs.com/acgoto/p/9483425.html