HDU 1002 A + B Problem II

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 303792    Accepted Submission(s): 58634

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
  
  

 

Author

Ignatius.L

 

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#include <stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<cmath>
#include <stdlib.h>

using namespace std;

string s1,s2;

string sum(string s1,string s2)
{
	if(s1.length()<s2.length())
	{
		string temp=s1;
		s1=s2;
		s2=temp;
	}
	int i,j;
	for(i=s1.length()-1,j=s2.length()-1;i>=0;i--,j--)
	{
		s1[i]=char(s1[i]+(j>=0?s2[j]-'0':0));
		if(s1[i]-'0'>=10)
		{
			s1[i]=char((s1[i]-'0')%10+'0');
			if(i) s1[i-1]++;
			else s1='1'+s1;
		}
	}
	return s1;
}


int main()
{
	int t1,t2,k=0;
	cin>>t1;
	t2=t1;
	while(t1--)
	{
		k++;
		cin>>s1>>s2;
		cout<<"Case "<<k<<":"<<endl;
		cout<<s1<<" + "<<s2<<" = "<<sum(s1,s2)<<endl;
		if(k<t2)cout<<endl;
		s1[0]=0,s2[0]=0;
	}
	return 0;
}