HDU1002 A + B problem II

Problem Description
I have a very simple problem for you. Given 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 both positive integers, A and B. Notice that the integers is very large, that M EANs should not process them by using 32-bit integer. Assume the length of each integer would not exceed 1000.

Output
For the test case, you should output of the lines. The first line was "Case #:", # means the number of the the test case. The second line is the equation "A + b = sum", sum means the result of A + B. Note there is some spaces int the Equati On. Output a blank line between the test cases.

Sample Input
2
1 2
112233445566778899 998877665544332211

Sample Output
Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

Author
Ignatius.l

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Test instructions: Sum of two large integers
Idea: It is in accordance with mathematical methods, from the single-digit beginning to add, if the full 10 into a, simulated childhood play with the Abacus to do add and subtract.

Where CE once, found that the operation of a string to use the string header, Ah, and CString is not a header file said.

#include <iostream> #include <cstdio> #include <cstring> #include <string> using namespace std;
    int main () {string A, B;
    int na[8000],nb[8000],sum[8000],pre;
    int num;
    CIN >> Num;
        for (int j = 1;j<=num;j++) {memset (sum,0,sizeof (sum));
        memset (na,0,sizeof (NA));
        memset (nb,0,sizeof (NB));
        Cin >> a >> b;
        Pre = 0;
        int Lena = A.length ();
        int lenb = B.length ();
        for (int i = 0;i<lena;i++) na[lena-i-1] = a[i]-' 0 ';
        for (int i = 0;i<lenb;i++) nb[lenb-i-1] = b[i]-' 0 '; int Lenx = Lena > LenB?
        Lena:lenb;
            for (int i = 0;i < Lenx; i++) {sum[i] = Na[i] + nb[i] + pre;
            Pre = 0;
                if (sum[i]>=) {pre = 1;
            Sum[i] = sum[i]%10;
        }} cout << "case" << J << ":" <<endl; cout <<
        A << "+" << b << "=";
        for (int i = lenx-1;i>=0;i--) cout<<sum[i];
        cout<<endl;
    if (j!=num) cout<<endl;
} return 0; }