#include <iostream>
using namespace std;
#define MAX 200
unsigned int num[MAX+10];
unsigned int num1[MAX+10];
int main()
{
string str1,str2;
cin>>str1>>str2;
memset(num,0,sizeof(num));
memset(num1,0,sizeof(num1));
int len = str1.length();int index = 0;
for (int i=len-1;i>=0;i--)
{
num[index++] = str1[i]-'0';
}
len = str2.length();index = 0;
for (int i=len-1;i>=0;i--)
{
num1[index++] = str2[i]-'0';
}
len = str1.length()>str2.length() ? str1.length() : str2.length();
for (int i=0;i<len;i++)
{
num[i] += num1[i];
if (num[i]>=10)
{
num[i] -= 10;
num[i+1]++;
}
}
bool flag = false;
for (int i=len;i>=0;i--)
{
if (flag||num[i])
{
flag = true;
cout<<num[i];
}
}
if (!flag)
{
cout<<"0";
}
return 0;
}
{//计算a1与a2的和,结果可在到c中,并返回c的长度
int i;
int len;
int k=0;//进位
if(l1<l2)len=l1;
else len=l2;
for(i=0;i<len;i++)
{
c[i]=(a1[i]+a2[i]+k)%10;
k=(a1[i]+a2[i]+k)/10;
}
if(l1>len)
{
for(i=len;i<l1;i++)
{
c[i]=(a1[i]+k)%10;
k=(a1[i]+k)/10;
}
if(k!=0)
{
c[l1]=k;
len=l1+1;
}
else len=l1;
}
else
{
for(i=len;i<l2;i++)
{
c[i]=(a2[i]+k)%10;
k=(a2[i]+k)/10;
}
if(k!=0)
{
c[l2]=k;
len=l2+1;
}
else len=l2;
}
return len;