hdu 1002

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 134859    Accepted Submission(s): 25637

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>
#define MAX 1000
using namespace std;
int main()
{
    int i,j,T,cont = 0;
    int A[MAX] = {0},B[MAX] = {0};
    char a[MAX],b[MAX];
    cin >> T;
    while (T--)
    {
        cin >> a >> b;
        for (i = 0; i < MAX; i++)
        {
            A[i] = 0;
            B[i] = 0;
        }
        int alen = strlen(a);
        int blen = strlen(b);
        for (i = 0; i < alen; i++)
        {
            A[i] = a[alen - 1 - i] - 48;
        }
        for (i = 0; i < blen; i++)
        {
            B[i] = b[blen - 1 - i] - 48;
        }
        int c = 0;//
        for (i = 0; i < MAX; i++)
        {
            int s = A[i] + B[i] + c;
            A[i] =  s % 10;
            c = s / 10;
        }
        cout<<"Case "<<++cont<<":"<<endl;
        for (i = 0; i < alen; i++)
        {
            cout << a[i];
        }
        cout<<" + ";
        for (i = 0; i< blen; i++)
        {
            cout << b[i];
        }
        cout<<" = ";
        for (i = MAX - 1; i >= 0; i--)
        {
            if (A[i])
                break;
        }
        for (j = i; j >= 0 ;j--)
        {
            cout << A[j];
        }
        cout << endl;
        if (T)
            cout << endl;
    }
    return 0;
}