codeforces1469D. Ceil Divisions

https://codeforces.com/contest/1469/problem/D

This observation found a better strategy

We choose a base c, and then let n/(n/c), (n/c)/(n/c/c)... and so on, turn these numbers into c, and then use c again at the end Turn them into 1, and finally use 2 to turn c into 1, then there will be 2 and 2 in the end, and the others are 1.

For other numbers that are not considered in the middle, we first let them all /n become 1

So for the numbers 3-n, only c must keep him /2, and then the numbers n/c and n/c/c must first change to c and then change to 1. The other numbers change to 1 at a time.

So this question is essentially to minimize log_c{n}+log_2{c}

People with a solid foundation in mathematics will do it right away. The uncle next door who is in his 20s who is still doing basic mathematics and science has been studying for a long time, but don’t learn from me.

Finally, try to find that c=8 is feasible under n=2e5

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int maxl=3e5+10;

int n,m,k,cnt,tot,cas;
int a[maxl];
bool vis[maxl];
char s[maxl];
typedef pair<int,int> p;
vector<p> ans,ans2;

inline void prework()
{
	scanf("%d",&n);
//	n=2e5;
	for(int i=1;i<=n;i++)
		vis[i]=false;
	ans.clear();
	int c=8,now=n,nxt;cnt=0;
	if(n<c)
	{
		for(int i=3;i<n;i++)
			ans.push_back({i,n});
		now=n;
		while(now>1)
			now=now/2+(int)(now%2!=0),ans.push_back({n,2});
		return;
	}
	while(1)
	{
		if(now<=c)
			break;
		vis[now]=true;
		a[++cnt]=now;
		nxt=now/c;
		if(now%c!=0)
			nxt+=1;
		now=nxt;
	}
	a[++cnt]=c;
	vis[c]=true;
	for(int i=3;i<n;i++)
	if(!vis[i])
		ans.push_back({i,n});
	for(int i=1;i<cnt;i++)
		ans.push_back({a[i],a[i+1]});
	for(int i=1;i<cnt;i++)
		ans.push_back({a[i],c});
	now=c;
	while(now>1)
		now=now/2+(int)(now%2!=0),ans.push_back({c,2});
}

inline void mainwork()
{
	
}

inline void print()
{
	tot=ans.size();
	printf("%d\n",tot);
	for(p d:ans)
		printf("%d %d\n",d.first,d.second);
}

int main()
{
	int t=1;
	scanf("%d",&t);
	for(cas=1;cas<=t;cas++)
	{
		prework();
		mainwork();
		print();
	}
	return 0;
}