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): 147889    Accepted Submission(s): 27893

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

 

 

由于处理的数比较大超过了long long 型，所以要变为数组处理比较好！

 解题代码：

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
char a[10003], b[10003], c[10003];

void deal(int lena, int lenb)
{
    if (lena > lenb)
    {
          lenb --;
        for (int i = lena - 1; i >= 0; i--)
        {
            //cout << b[lenb]<<endl;
            if(lenb >= 0)
            {
                b[i] = b[lenb];
                b[lenb] = '0';
                lenb --;
            }
            else b[i] = '0';
        }
        return;
    }
    if (lena < lenb)
    {
        lena--;
        for (int i = lenb - 1; i >= 0; i --)
        {
            if (lena >= 0)
            {
                a[i] = a[lena];
                a[lena] = '0';
                lena --;
            }
            else a[i] = '0';
        }
        return;
    }
    return ;
}
int main()
{
    int n, num = 1;
    int lena, lenb;
    cin >> n;
    while (n--)
    {
        cin >> a >> b;
        lena = strlen (a);
        lenb = strlen (b);
        printf ("Case %d:\n", num++);
        cout << a <<" + " << b<<" = ";        
        deal(lena, lenb);
        int d = 0;
           int j = 0;
          for (int i = (lena > lenb ? lena : lenb) - 1; i >= 0; i--)
          {
                int temp = a[i] - 48 + b[i] - 48 + d;
                d = temp/10;
                c[j] = temp%10 + 48;
                j++;
           }
           c[j] = d;
           if (d)
           {
                c[j] = d + 48;
               c[j+1] = 0; 
           }
        for (int i =j-1; i >= 0; i --)
               cout << c[i];
          cout << endl;
        if (n)
            cout <<endl;
    }
    return 0;
}