First approach: Optimize the straightforward solution.

Multiply instead of divide. What happens to the last digit when multiplying?

Consider islands of two colours and the bridges between them.

Iterate over all possible combination, order and direction of extra edges.

First approach: 2D segment tree (or quadtree) with a set or vector on each node, representing the set of barriers that cover this node.

#include <bits/stdc++.h>
using namespace std;
#define Maxn 4000007
bool vis[Maxn];
int a[4007],b[4007],n;
int main()
{
	scanf("%d",&n);
	memset(vis,false,sizeof(vis));
	for (int i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);
		vis[a[i]]=true;
	}
	for (int i=1;i<=n;i++)
	{
		scanf("%d",&b[i]);
		vis[b[i]]=true;
	}
	int ans=0;
	for (int i=1;i<=n;i++)
		for (int j=1;j<=n;j++)
			if (vis[a[i]^b[j]]) ++ans;
	if (ans%2==0) printf("Karen\n"); else printf("Koyomi\n");
	return 0;
}

#include <stdio.h>

int main()
{
    puts("Karen");
    return 0;
}

#include <bits/stdc++.h>
using namespace std;
long long L,R;
int ans;
int main()
{
	scanf("%lld%lld",&L,&R);
	if (R-L>=10) printf("%d\n",0);
	else
	{
		ans=1;
		for (long long i=L+1;i<=R;i++)
			ans=(1LL*ans*(i%10))%10;
		printf("%d\n",ans);
	}
	return 0;
}