Description

The bored Three-God get another boring question.
This problem is ask you plus two big nubmer, please help him, when you solve this problem you can

speak to Three-God,“Oh, Please give me a diffculte one, this one is too easy”.

Input

There are muti-case
For each case, there are two integers, n, m (0 < n, m < 10^10000).

Output

Calculate two integers’ sum

Sample Input

1 1
2 3
10000 100000

Sample Output

2
5
110000
很明显，这是一道大数问题，但是此题比较坑的地方就是需要考虑前导0，比如01+01答案是02而不是2，其它的就是用大数加法的思路解决就行了，代码如下。

#include <iostream>
#include<string.h>
#include<algorithm>
#include<stdio.h>
using namespace std;
char v[100005],s[100005];
int  x[100005],y[100005],z[100005];
int main()
{


    while(scanf("%s",v)!=EOF)
    {
         memset(x,0,sizeof(x));
         memset(y,0,sizeof(y));
         memset(z,0,sizeof(z));
        scanf("%s",s);
        int a=strlen(v),b=strlen(s);
        int c=max(a,b);
        for(int i=a-1,j=0;i>=0;i--)
        {
            x[j++]=v[i]-'0';
        }
        for(int i=b-1,j=0;i>=0;i--)
        {
            y[j++]=s[i]-'0';
        }
        for(int i=0;i<c;i++)
        {
            z[i]+=x[i]+y[i];
            if(z[i]>=10)
            {
                z[i+1]=z[i]/10;
                z[i]=z[i]%10;
            }
        }
        if(z[c]!=0)
        {
            printf("%d",z[c]);
        }
        for(int i=c-1;i>=0;i--)
        {
            printf("%d",z[i]);
        }
        printf("\n");
    }
    return 0;
}