heidaizx — Wed, 21 May 2008 01:24:00 GMT

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14324 Accepted Submission(s): 2486

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<cstring>
using namespace std;
char *add(char *a,char *b)
{
     int i,j,k=0;
     int temp[1001];
     int length1 = strlen(a);
     int length2 = strlen(b);
     for(i = length1-1, j = length2 -1 ; i >= 0&&j >= 0; --i,--j)
           temp[k++] = a[i] + b[j] - '0' - '0';
     for(;i >= 0; --i)
          temp[k++] = a[i] - '0';
     for(;j >= 0; --j)
            temp[k++] = b[j] - '0';
     temp[k] = 0;
     for(i = 0;i < k; ++i)
     {
           temp[i+1] += temp[i] / 10;
           temp[i] %=10;
     }
     if(!temp[k]) k--;
     for(i = 0;i <= k;i++)
           a[i] = temp[k-i] + '0';
     a[k+1] = '\0';
     return a;
}
int main()
{
    char a[1001],b[1001],*r;
    int number = 0;
    int k = 1;
    cin >> number;
    while(number--)
    {
                   if(cin >> a >> b)
                   {
                          cout << "Case "<< k++ << ":\n" << a << " + " << b <<" = ";
                          r = add(a,b);
                          cout << r;
                          cout << endl;
                          if(number)
                                    cout<<endl;
                   }
    }
    return 0;
}

heidaizx 2008-05-21 09:24 发表评论

欧美精品123区,国产欧美一区在线,悠悠资源网久久精品

A + B Problem II