A+B Problem II

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.

輸入

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.

輸出

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.

樣例輸入

2
1 2
112233445566778899 998877665544332211

樣例輸出

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

這道題簡單一看只是大數定理，但其實比較坑爹的是  輸入00008+2應該輸出10，需要過濾前導0，這道題如果不是前導0我只用了半小時就做對了，還去拉了個大便，結果看討論的說有前導零，又費了一個小時，唉！！

#include<iostream>
#include<string>
#include<cstdio>
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<lenb||i<lena;i++)
		{
			h=a[i]+b[i]+k;
			f[i]=h%10;
			k=h/10;
		}
		if(k!=0)
		    f[i++]=k; 
		cout<<"Case "<<y<<":"<<endl<< a1 <<" + "<< b1 <<" = ";
		p=0;
		for(j=i-1;j>=0;j--)
		{
		 	if(p==0&&f[j]==0)
		 	{
		 	    continue;
		 	}
		 	else
		 	{
		 	    p=1;
		 	    cout<<f[j];
		 	}
		}
		cout<<endl;
	}
	return 0;
}