1002 A + B Problem II [ACM刷题]

这一段时间一直都在刷OJ，这里建一个博客合集，用以记录和分享算法学习的进程。

github传送门：https://github.com/haoyuanliu/Online_Judge/tree/master/HangDianOJ

 

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 CODE

//这一版AC代码，由于题目对于给定的输入输出太多限定，所以代码有很多冗余的部分，
//核心思想就是使用堆栈的方法，使用三个堆栈实现了大数的加法
#include <iostream>
#include <string>
#include <stack>
#include <vector>
using namespace std;
int count = 0;//声明一个count全局变量，用作进位计算
int main()
{
    int i, n;
    vector<string> vec1;
    vector<string> vec2;
    string str1, str2;
    stack<char> stk1;
    stack<char> stk2;
    stack<int> rstk;
    cin >> n;
    //完成大数字符串的输入
    for(i = 0; i < n; ++i)
    {
        cin >> str1 >> str2;
        vec1.push_back(str1);
        vec2.push_back(str2);
    }
    for(i = 0; i < n; ++i)
    {
        //将数字以char字符形式读入并push进堆栈中
        for(const auto x : vec1[i])
        {
            stk1.push(x);
        }
        for(const auto y : vec2[i])
        {
            stk2.push(y);
        }
        //取堆栈元素进行加法运算
        while(!stk1.empty() && !stk2.empty())
        {
            int temp = stk1.top() - '0' + stk2.top()  - '0' + count;
            if(temp > 9)
            {
                temp %= 10;
                count = 1;
            }
            else
            {
                count = 0;
            }
            rstk.push(temp);
            stk1.pop();
            stk2.pop();
        }
        //进行多余高位的push
        while(!stk1.empty())
        {
            int temp = stk1.top() - '0' + count;
            if(temp > 9)
            {
                temp %= 10;
                count = 1;
            }
            else
            {
                count = 0;
            }
            rstk.push(temp);
            stk1.pop();
        }
        while(!stk2.empty())
        {
            int temp =  stk2.top() - '0' + count;
            if(temp > 9)
            {
                temp %= 10;
                count = 1;
            }
            else
            {
                count = 0;
            }
            rstk.push(temp);
            stk2.pop();
        }
        while(count == 1)
        {
            rstk.push(1);
            count = 0;
        }
        cout << "Case " << i+1 << ":" << endl;
        cout << vec1[i] << " " << "+" << " " << vec2[i] << " = ";
        //最终结果打印出来即可
        while(!rstk.empty())
        {
            cout << rstk.top();
            rstk.pop();
        }
        cout << endl;
        while(i < n-1)
        {
            cout << endl;
            break;
        }
    }
    return 0;
}

 精简代码

/*这一版是简化版的大数加法代码
 *去除了一部分输入输出规则，以核心代码为主
 */
#include <iostream>
#include <string>
#include <stack>
#include <vector>
using namespace std;
int count = 0;//声明一个count全局变量，用作进位计算
int main()
{
    int i = 0;
    string str1, str2;
    stack<char> stk1;
    stack<char> stk2;
    stack<int> rstk;
    while(cin >> str1 >> str2)
    {
        //将数字以char字符形式读入并push进堆栈中
       for(const auto x : str1)
        {
            stk1.push(x);
        }
        for(const auto y : str2)
        {
            stk2.push(y);
        } 

         //取堆栈元素进行加法运算
        while(!stk1.empty() && !stk2.empty())
        {
            int temp = stk1.top() - '0' + stk2.top()  - '0' + count;
            if(temp > 9)
            {
                temp %= 10;
                count = 1;
            }
            else
            {
                count = 0;
            }
            rstk.push(temp);
            stk1.pop();
            stk2.pop();
        }
        //进行多余高位的push
        while(!stk1.empty())
        {
            int temp = stk1.top() - '0' + count;
            if(temp > 9)
            {
                temp %= 10;
                count = 1;
            }
            else
            {
                count = 0;
            }
            rstk.push(temp);
            stk1.pop();
        }
        while(!stk2.empty())
        {
            int temp =  stk2.top() - '0' + count;
            if(temp > 9)
            {
                temp %= 10;
                count = 1;
            }
            else
            {
                count = 0;
            }
            rstk.push(temp);
            stk2.pop();
        }
        while(count == 1)
        {
            rstk.push(1);
            count = 0;
        }
        cout << "Case " << i+1 << ":" << endl;
        cout << str1 << " " << "+" << " " << str2 << " = ";
        //最终结果打印出来即可
        while(!rstk.empty())
        {
            cout << rstk.top();
            rstk.pop();
        }
        cout << endl << endl;
        ++i;
    }
    return 0;
}
    

PS：网上OJ的输入输出的形式规范实在是太蛋疼了！！！