Bored Three-God

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J - Bored Three-God

NBUT - 1228

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

Hint

无

题意：大数A+B 这题有个很坑的点就是输出前导零列如 005 006 输出 011

#include<stdio.h>
#include<string.h>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
#include<math.h>
#include<queue>
#include <deque>
#include <cstdio>
#include<list>
#define LL long long
using namespace std;
char a[1001000];
char b[1001000];
int a1[1001000];
int b1[1001000];
int c[1001000];
int main()
{
while(~scanf("%s%s",a,b))
{
memset(a1,0,sizeof(a1));
memset(b1,0,sizeof(b1));
memset(c,0,sizeof(c));
int k1=strlen(a);
int k2=strlen(b);
for(int i=0;i<k1;i++)
{
a1[i]=a[k1-1-i]-'0';
}
for(int i=0;i<k2;i++)
{
b1[i]=b[k2-1-i]-'0';
}
int k=max(k1,k2);
int d=0;
for(int i=0;i<=k;i++)
{
int t=a1[i]+b1[i]+d;
c[i]=t%10;
d=t/10;
}
// int l=-1;
// for(int i=k;i>=0;i--)
// {
// if(c[i]!=0)
// {
// l=i;
// break;
// }
// }
if(c[k]!=0)
printf("%d",c[k]);
for(int i=k-1;i>=0;i--)
{
printf("%d",c[i]);
}
printf("\n");
}
}