go play fortnite u fortnite kid

you should just print a + b

a, b = map(int, input().split())
print(a + b)

But if u solvin' extreme a+b with really big a and b, and few memory, the only way is to write long arythmetics on C++. Maybe long arythmetics in Java will work too (they are already in Java).

if $$$a+b\le10^8$$$:

ans = 0;
for(int i = 1; i <= a; i++)
  ans++;
for(int i = 1; i <= b; i++)
  ans++;

otherwise I dont know

This joke is too old

std::cout << ( ( 2 * a + 2 * b ) — ( a + b ) ) + ( __gcd( (int)1e7 , 1 ) — pow( 2 , 0 ) ) ;

xD

Auto comment: topic has been updated by plagues (previous revision, new revision, compare).

What are the min and max values for a and b? Let's see if we can write just some if-statements

А шо все такие англичане?

hey guys, i found super simple solution with complexity $$$O(\log ab)$$$

#include <iostream>

using namespace std;

long long ans = 0;

void add_to_ans(long long x) {
	long long tmp = 1;
	while (x) {
		if (x & 1) {
			ans += tmp;
		}
		x >>= 1;
		tmp <<= 1;
	}
}

int main()
{
	long long a, b;
	cin >> a >> b;
	add_to_ans(a);
	add_to_ans(b);
	cout << ans << "\n";

	return 0;
}

Hey, guys. I found a faster solution!

#include <bits/stdc++.h>

using namespace std;

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int a, b;
    cin >> a >> b;
    int ans = 0;
    for (int i = 0; i < a + b; i++) {
        ans++;
    }
    cout << ans;
    return 0;
}