[Codeforces Round #666 (Div. 2)] B. Power Sequence brute force enumeration

Topic link: B. Power Sequence

Title

Give you a sequence a of length n, let you make it into a sequence called "power sequence" through some operations-there is a c such that (0≤i≤n-1), $a_i=c^i$。

1. You can change the position of each element in the sequence. This operation is not counted in the operand.
2. You can choose any element to increase or decrease by 1. Each time this operation is performed, the operand is +1.

Ask the minimum number of operands required to make it "power squence".

answer

Let’s analyze this question first, $a_i\le 1e9，n\le 1e5$ .
Then we can draw a conclusion that the operand cannot exceed$1e14$ , because when$a_{i}Allare1e9$ , n is$At\; 1e5$ , we can get the minimum value by taking c=1.

Then we continue to analyze
when c = $10^4$ o’clock, then the power sequence sequence is$1，10^4,10^8,10^{.\; 12}$
When = C$10^3$ o'clock, then the power sequence sequence is$1，10^3,10^6,10^9,10^{12}$

It is easy to find that when c>=1000, n cannot exceed 4. If it exceeds 4, we can get the answer when we take c=1.

So for this question we can enumerate c. Since $a_i\le 1e9,n\ge 3$ , so$c\le \sqrt{1e9}$.
In calculating the ai {a_i} of the power sequence$a_i$, If $a_i\ge 1e14$ , then exit the loop.

Due to pruning, the time complexity will not exceed.

Code

#include<iostream>
#include <sstream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
using namespace std;
//extern "C"{void *__dso_handle=0;}
typedef long long ll;
typedef long double ld;
//typedef int fuck;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define pii pair<int,int>
#define lowbit(x) x&-x
//#define int long long

const double PI=acos(-1.0);
const double eps=1e-6;
//const ll mod=1e9+7;
const ll inf=1e18;
const int maxn=1e5+10;
const int maxm=100+10;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

ll a[maxn];
int main()
{
    
    
	int n; 
	scanf("%d",&n);
	for(int i=0;i<n;i++) scanf("%lld",&a[i]);
	sort(a,a+n);
	ll ans=inf;
	for(ll c=1;c<=sqrt(1e9)+1;c++)
	{
    
    
		ll tmp=0,tt=1;
		int flag=1;
		for(int i=0;i<n;i++)
		{
    
    
			if(tt>=1e14) {
    
     flag=0;break;}
			tmp+=abs(a[i]-tt);
			tt*=c;
		}
		if(flag) ans=min(ans,tmp);
	}
	printf("%lld\n",ans);
}