1002 A + B Problem II

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A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 526070    Accepted Submission(s): 100589

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

 

AC代码：

#include<iostream>
#include<string>
#include<vector>
using namespace std;
int main()
{
    int i, n, j, k;
    vector<char> a;
    string s1, s2, s;
    while (cin >> n) {
        for (i = 0; i < n; i++) {
            a.clear();
            cin >> s1 >> s2;
            j = (int)s1.size() - 1;
            k = (int)s2.size() - 1;
            while (j >=0 && k >= 0 ) {
                a.push_back(s1[j] - '0' + s2[k] - '0');
                j--;
                k--;
            }
            while (j >= 0) {
                a.push_back(s1[j] - '0');
                j--;
            }
            while (k >= 0) {
                a.push_back(s2[k] - '0');
                k--;
            }
            for (j = 0; j < (int)a.size() - 1; j++) {
                if (a[j] >= 10) {
                    a[j+1] += a[j] / 10;
                    a[j] %= 10;
                }
            }
            if (a[a.size() -1 ] >= 10 ) {
                int t = a[a.size() - 1] / 10;
                a[a.size() - 1] = a[a.size() - 1] % 10;
                a.push_back(t);
            }
            s.clear();
            for (j = (int)a.size() - 1; j >=0 ; j--) {
                s.push_back(a[j] + '0');
            }
            cout << "Case " << i + 1 << ":" << endl;
            if (i == n - 1) {
                cout << s1 << " + " << s2 << " = " << s << endl;
            }
            else {
                cout << s1 << " + " << s2 << " = " << s << endl << endl;
            }
            
        }
    }
    return 0;
}