【OJ】 大數加法問題 A + B Problem II

1032: A + B Problem II

題目描述

I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

輸入

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.

輸出

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 in the equation. Output a blank line between two adjacent test cases.

樣例輸入

4
1 2
112233445566778899 998877665544332211
998 2
4568 0012 

樣例輸出

Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

Case 3:
998 + 2 = 1000

Case 4:
4568 + 0012 = 4580

解題代碼

問題分析：簡單大數加法問題，但是這個題目裏面有一個坑，就是開頭爲0這種情況，解決方法也很簡單，開頭就是0的話，不入棧。

#include <iostream>
#include <stack>
#include <cstring>
using namespace std;

/**
 *  大數加法 written by Smileyan
 *  需要注意字符串開頭的0這個問題
 */
stack<int> BigNumAdd(char *str1,char *str2)
{
    int i;
    stack<int> stack1,stack2;
    stack<int> result;

    // 專門用來解決開頭的 0
    i=0;
    while(str1[i] == '0')     i++;

    for(; str1[i] != '\0'; i++)
    {
        stack1.push(str1[i]-'0');
    }

    i=0;
    while(str2[i] == '0')     i++;
    for(; str2[i] != '\0'; i++)
    {
        stack2.push(str2[i]-'0');
    }

    // 用來記錄是否需要進位
    bool flag=false;

    while(stack1.empty()==false || stack2.empty() == false || flag)
    {
        int x1 = 0;
        if(stack1.empty()==false)
        {
            x1 = stack1.top();
            stack1.pop();
        }
        int x2 = 0;
        if(stack2.empty()==false)
        {
            x2 = stack2.top();
            stack2.pop();
        }

        // 如果兩個都爲空，則就是 0+0+進位
        int r = x1+x2+flag;
        if(r >= 10)
        {
            r -= 10;
            flag = true;
        }
        else
        {
            flag = false;
        }
        result.push(r);
    }
    return result;
}


// written by Smileyan 多謝關注
int main()
{
    char str1[250],str2[250];
    int n;
    stack<int> result;
    int i;
    cin>>n;
    for(i=1; i<=n; i++)
    {
        cin>>str1>>str2;
        result = BigNumAdd(str1,str2);
        cout<<"Case "<<i<<":"<<endl;
        cout<<str1<<" + "<<str2<<" = ";

        while(result.empty()==false)
        {
            int num=result.top();
            result.pop();
            cout<<num;
        }
        cout<<endl<<endl;
    }
    return 0;
}

總結

和大數減法 差不多，唯一的區別就是題目中添加了開頭就是0這種 現實生活中不可能出現的情況。

Smileyan
2019年9月21日 10:16