@[weak_in_code](/user/753640) 不太理解,时间复杂度应该不含常数吧?所以应该是 $O(n\sqrt{n}\log n \log n)$ 更小?
by szh_DNCB @ 2024-04-08 20:22:56
```cpp
#include<bits/stdc++.h>
using namespace std;
int l,r;
int main()
{
for(int i=0;i<=100000;i++)
{
double A=10.0*i*sqrt(i)*log2(i)*log2(i);
double B=i*i*1.0*log2(i)*1.0*log2(i);
// cout<<A<<" "<<B<<"\n";
if(A>B) r++;
else l++;
}
printf("%s",l>r?"left":"right");
return 0;
}
```
@[weak_in_code](/user/753640)
by z_z_b_ @ 2024-04-08 20:34:50
@[weak_in_code](/user/753640) 第一种,sqrt(n)如果小于10的话两种也都无所谓了
by NightDiver @ 2024-04-08 20:35:13
@[NightDiver](/user/985711) @[z_z_b_](/user/956129) @[szh_DNCB](/user/1073754) thx,主要是怕卡不掉 $O(n^2\log n\log n)$ 的做法。
此帖结。
by weak_in_code @ 2024-04-08 20:38:18