A + B

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 input

Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

思路：使用字符串儲存兩個大數據，再用一個數組c來儲存他們的和，用變量p,q分別指向他們的個位數，因為是從左到右輸入的，所以是從strlen(a) - 1開始，將其個位數相加，使用sum儲存，sum%10儲存在c中sum/10補在下一位，一直循環到p和q等於0，註意當其中一個為0時停止其對應相加。

#include <iostream>
#include 
 
<cstring>
#include <cstdio>
using namespace std;
char a[1005], b[1005];
int c[1005];
int main()
{
    int t;
    cin >> t;
    for(int i = 1; i <= t; i++)
    {
        memset(a, ‘0‘, sizeof(a));
        memset(b, ‘0‘, sizeof(b));
        memset(c, 0, sizeof(c));
        getchar();
        cin >> a >> b;
        int sum = 0;
        int p = strlen(a) - 1, q = strlen(b) - 1;
        int k = 0;
        while(1)
        {
            if(p >= 0 && q >= 0)
            sum = a[p] - 48 + b[q] - 48 + sum;
            if(p < 0 && q >= 0)
            sum =  + b[q] - 48 + sum;
            if(p >= 0 && q < 0)
            sum = a[p] - 48 +  + sum;
            c[k++] = sum%10;
            sum = sum/10;
            if(p < 0 && q < 0)break;
            p--;
            q--;
        }
        cout << "Case " << i << ‘:‘;
        cout << endl;
        for(int j = 0; j < strlen(a); j++)
        cout << a[j];
        cout << " + ";
        for(int j = 0;j < strlen(b); j++)
        cout << b[j];
        cout << " = ";
        for(int j = k - 2; j >= 0; j--)
        cout << c[j];
        cout << endl;
        if(i != t)cout << endl;

    }
    return 0;
}

大整數加法 A + B