大数 a+b

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

 

#include<iostream>
#include<string>
using namespace std;

#define P(x) \
cout << #x " " << x << ":" << endl;

#define PE(a, b, c, sum) \
cout << a << " + " <<  b << " = " << (c ? "1" : "") << sum << endl;
int main()
{
    int T;
    cin >> T;
    int Case = 1;
    while(T--)
    {
        P(Case);
        ++Case;
        string a, b, sum;
        cin >> a >> b;
        //cout << a << " + " <<  b << " = " ;
        int l_a = a.size()-1, l_b = b.size()-1, c = 0;
        sum = (l_a > l_b ? a : b);
        int i, j;
        for(i = l_a, j = l_b; i >= 0 && j >= 0; --i, --j)
        {
            int s  = a[i] - '0' + b[j] - '0' + c;
            sum[i>j?i:j] = s % 10 + 48;
            c = s / 10;
        }
        while(i >= 0)
        {//处理进位
            int s = sum[i] - '0' + c;
            sum[i] = s % 10 + 48;
            c = s / 10;
            i--;
        }
        while(j >= 0)
        {//处理 进位
            int s = sum[j] - '0' + c;
            sum[j] = s % 10 + 48;
            c = s / 10;
            j--;
        }
        // cout << (c ? "1" : "") << sum << endl;
        PE(a, b, c, sum);
        if(T)
            cout << endl;
    }
    return 0;
}

#include<iostream>
#include<string>
using namespace std;

const int maxn = 500;
#define P(x) \
cout << #x ": " << x << endl;
int main()
{
    string a;
    cin >> a;
    //string str;
    //string str("shi");
    string str(maxn, 'a');
    //str = ""; //长度也变为了0
    for(int i = 0; i != a.size(); ++i)  //the same above
    //for(int i = a.size()-1; i >= 0; --i)
        str[i] = a[i];
    P(a);
    P(str);
    return 0;
}

/*
str: ""
output: ""

str:"shi"
output: san

由此可见，string 对象是有默认长度的(根据其赋值对象的长度决定)
与字符数组类似的，只不过不需明确指明，可根据赋值对象动态的适应并
改变它。字符赋值，其实质是在原有的基础上修改，是有长度限制的。
但是后面以后str = "";以后长度也都变为了0.由此可见，使用字符赋值的
关键是如何确定他的可能长度。
*/