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[1000],b[1000],res[1001];
char cha(char a);

void add();

int main()
{
    int t;
    cin>>t;
    for(int i=1;i<=t;i++)
    {
        cin>>a>>b;
        cout<<"Case "<<i<<":"<<endl<<a<<" + "<<b<<" = ";
        add();
        cout<<res<<endl;
        if(i==t) break;
        cout<<endl;
        memset(res,0,sizeof(res));
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
    }
    return 0;
}

void add()
{
    int len1,len2,len,i,j;
    len1=strlen(a);
    len2=strlen(b);
    len=max(len1,len2);
    strrev(a);
    strrev(b);
    for(i=0;i<len;i++)
    {
        char t=cha(res[i]);
        res[i]=(char)((t+(cha(a[i])+cha(b[i]))%10)%10+'0');
        res[i+1]=(t+cha(a[i])+cha(b[i]))/10+'0';
    }
    if(res[len]=='0') res[len]='\0';
    strrev(res);
    if(res[0]=='0')
    {
        for(i=0;res[0]=='0';i++)
        {   
            for(j=0;j<len;j++)
            {
                res[j]=res[j+1];
            }
        }
    }
}

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

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大数字加法（hduoj）

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