ps:正好有朋友问这题，就写了个超多注释的超详细版代码。每步都有解释。希望也能帮到其他人。

问题：

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 434632    Accepted Submission(s): 84594

 

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

 

代码：

#include<iostream>
#include<string>
#include<cstdio>
using namespace std;
int i,j,y,n,k,h,p,lena,lenb;//先定义一批变量，用于以后存数 
int a[1010]={0},b[1010]={0},sum[1010]={0},f[1010]={0};//存大数位,加和,全部遍历为0 ，为了让之后的判断更方便 
int main()
{
	string a1,b1;//定义两个字符串先把数输入进来 
	cin>>n;//输入n组数据 
	for(y=1;y<=n;y++)
	{
		cin>>a1>>b1;//输入大数 
		lena=a1.length();//看两个大数的数位 
		lenb=b1.length();
		//a1，b1字符串中的字符a转化为数 
		for(i=lena-1;i>=0;i--)
		    a[lena-1-i]=a1[i]-'0';
		for(i=lenb-1;i>=0;i--)
		    b[lenb-1-i]=b1[i]-'0';
		//用k来判断a+b会不会超过10  进位 
	    k=0;
	    //进行逐位a+b，用f【】存放 
		for(i=0;i<lenb||i<lena;i++)
		{
			h=a[i]+b[i]+k;
			f[i]=h%10;
			k=h/10;
		}
		//进行一个判断看是否最后一味相加也进位 
		if(k!=0)
		    f[i++]=k; 
		cout<<"Case "<<y<<":"<<endl<< a1 <<" + "<< b1 <<" = ";
		for(j=i-1;j>=0;j--)
		{
		 	cout<<f[j];
		}

		cout<<endl<<endl;
	}
	return 0;
}