HDOJ 1002

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 124589    Accepted Submission(s): 23993

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

 

 

#include <iostream>
using namespace std;

void sum(char (&a)[1001],char (&b)[1001]);
int main()
{
    int T, count_T;
    char a[1001], b[1001];
    cin >> T;
    for(count_T=1; count_T <= T; count_T++) 
    {
        cin >> a >> b;

        if(count_T==1) 
        {
            cout << "Case " << count_T << ":" << endl;
        }
        else 
        {
            cout << "\nCase " << count_T << ":" << endl;
        }

        cout << a << " + " << b << " = ";

        sum(a,b);
    }
    return(0);
}
void sum(char (&a)[1001],char (&b)[1001])
{
    char c[1002];    
    int la, lb;
    la=strlen(a)-1;
    lb=strlen(b)-1;
    int i;
    int flag=0;
    for(i=0;la>=0 && lb>=0;la--,lb--,i++)
    {
        flag+=(a[la]-'0')+(b[lb]-'0');
        c[i]=flag%10;
        flag/=10;
    }
    for(la=la;la>=0;la--,i++)
    {
        flag+=(a[la]-'0');
        c[i]=flag%10;
        flag/=10;
    }
    for(lb=lb;lb>=0;lb--,i++)
    {
        flag+=(b[lb]-'0');
        c[i]=flag%10;
        flag/=10;
    }
    if(flag)
    {
        c[i]=flag;
        i++;
    }
    for(i--;i>=0;i--)
    {
        cout<<(char)(c[i]+'0');
    }
    cout<<endl;
}