2016年：杭电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): 295003    Accepted Submission(s): 56796

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

 

#include <iostream>
#include <string>
#include <stdio.h>
#include <stdlib.h>
#include <cstring>
using namespace std;

int main()
{
    int n,l=1,r;
    cin>>n;
    r=n;
    while(n--)
    {
        string a,b;
        cin>>a>>b;
        int len1=a.length();
        int len2=b.length();
        int m[1001],n[1001],q[1001];
        memset(m,0,sizeof(m));
        memset(n,0,sizeof(n));
        memset(q,0,sizeof(q));
        int x=0,y=0;
        for(int i=len1-1; i>=0; i--)
        {
            m[i]=a[x]-48;
            x++;
        }
        for(int i=len2-1; i>=0; i--)
        {
            n[i]=b[y]-48;
            y++;
        }
        int max;
        if(len1>len2)
        {
            max=len1;
        }
        else
        {
            max=len2;
        }
        cout<<"Case "<<l<<":"<<endl;
        for(int i=len1-1; i>=0; i--)
        {
            cout<<m[i];
        }
        cout<<" + ";
        for(int i=len2-1; i>=0; i--)
        {
            cout<<n[i];
        }
        cout<<" = ";
        for(int i=0; i<max+1; i++)
        {
            m[i]=m[i]+n[i];

            if(m[i]>9)
            {
                m[i]=m[i]-10;
                m[i+1]=1+m[i+1];
            }
        }

        if(m[max-1]>=0)
            cout<<m[max-1];
        for(int i=max-2; i>=0; i--)
        {

            cout<<m[i];
        }

        cout<<endl;
        if(l<r)
        {
            cout<<endl;
        }
        l++;
    }
    return 0;
}