A+B Problem II——大数定理

A+B Problem II

时间限制： 3000 ms  |  内存限制： 65535 KB

难度： 3

描述

I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

A,B must be positive.

输入

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.

输出

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.

样例输入

2
1 2
112233445566778899 998877665544332211

样例输出

Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

这道题简单一看只是大数定理，但其实比较坑爹的是  输入00008+2应该输出10，需要过滤前导0，这道题如果不是前导0我只用了半小时就做对了，还去拉了个大便，结果看讨论的说有前导零，又费了一个小时，唉！！

#include
#include
#include
using namespace std;
int i,j,y,n,k,h,p,lena,lenb;
int a[1000],b[1000],sum[1000],f[1000];
int main()
{
	string a1,b1;
	cin>>n;
	for(y=1;y<=n;y++)
	{
		cin>>a1>>b1;
		lena=a1.length();
		lenb=b1.length();
		for(i=0;i<1000;i++)
		{
			a[i]=0;b[i]=0;f[i]=0;
		}
		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';
	    k=0;
		for(i=0;i