大数字加法（hduoj）

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

 

Author

Ignatius.L

---

这是一道简单的加法题，但是数字特别大，就连long long都装不下那种~

需要用字符串来装输入数据

再按照类似小学列竖式的方法来计算……

因为竖式是从右往左，所以先把a，b字符串反转，计算完以后再反转回去~

---

#include<iostream>
#include<cstring>

using namespace std;

char a[],b[],res[];
char cha(char a);

void add();

int main()
{
    int t;
    cin>>t;
    for(int i=;i<=t;i++)
    {
        cin>>a>>b;
        cout<<"Case "<<i<<":"<<endl<<a<<" + "<<b<<" = ";
        add();
        cout<<res<<endl;
        if(i==t) break;
        cout<<endl;
        memset(res,,sizeof(res));
        memset(a,,sizeof(a));
        memset(b,,sizeof(b));
    }
    return ;
}

void add()
{
    int len1,len2,len,i,j;
    len1=strlen(a);
    len2=strlen(b);
    len=max(len1,len2);
    strrev(a);
    strrev(b);
    for(i=;i<len;i++)
    {
        char t=cha(res[i]);
        res[i]=(char)((t+(cha(a[i])+cha(b[i]))%)%+'');
        res[i+]=(t+cha(a[i])+cha(b[i]))/+'';
    }
    if(res[len]=='') res[len]='\0';
    strrev(res);
    if(res[]=='')
    {
        for(i=;res[]=='';i++)
        {
            for(j=;j<len;j++)
            {
                res[j]=res[j+];
            }
        }
    }
}

char cha(char a)
{
    if(a!='\0') return a-'';
    else return '\0';
}