题目链接：https://cn.vjudge.net/problem/NBUT-1228

题目

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

就是一个大数加法；
坑点就是要输出前导0；
比如005， 006要输出011。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
int a[100003],b[100003],c[100004];
char x[100005],y[100005];
int main()
{
    while(~scanf("%s%s",x,y))
    {
       memset(a,0,sizeof(a));
       memset(b,0,sizeof(b));
       memset(c,0,sizeof(c));
        int la=0;
        int lb=0;
        int lc=0;
        int lx=strlen(x);
        int ly=strlen(y); int w=0;int k=0;
        for(int i=lx-1;i>=0;i--)
        {

            a[w++]=x[i]-'0';
        }
        for(int i=ly-1;i>=0;i--)
        {

            b[k++]=y[i]-'0';
        }
        int mm=max(w,k);

        for(int i=0;i<mm;i++)
        {
            int s=a[i]+b[i];

            c[i]+=s;

            c[i+1]=c[i]/10;
             c[i]%=10;
        }
        int flog=0;
        if(c[mm])
        {
            printf("%d",c[mm]);
        }
        for(int j=mm-1;j>=0;j--)
        {
                printf("%d",c[j]);

        }
        printf("\n");
    }
    return 0;
}