[CodeForces-1397B] [Problem B]

Let’s call a list of positive integers a0,a1,…,an−1 a power sequence if there is a positive integer c, so that for every 0≤i≤n−1 then ai=ci.

Given a list of n positive integers a0,a1,…,an−1, you are allowed to:

Reorder the list (i.e. pick a permutation p of {0,1,…,n−1} and change ai to api), then
Do the following operation any number of times: pick an index i and change ai to ai−1 or ai+1 (i.e. increment or decrement ai by 1) with a cost of 1.
Find the minimum cost to transform a0,a1,…,an−1 into a power sequence.

Input
The first line contains an integer n (3≤n≤105).

The second line contains n integers a0,a1,…,an−1 (1≤ai≤109).

Output
Print the minimum cost to transform a0,a1,…,an−1 into a power sequence.

#include <bits/stdc++.h>
#define ll long long

using namespace std;

const int maxn = 1e5+3;
ll a[maxn];
int main()
{
    
    
    ios::sync_with_stdio(false);
    int n,m=1;
    cin>>n;
    for(int i=1;i<=n;i++)
        cin>>a[i];
    sort(a+1,a+n+1);
    ll res=1e18;
    for(ll c=1;;c++)
    {
    
    
        ll sum=0,now=1;
        for(ll i=1;i<=n;i++)
        {
    
    
            sum+=abs(a[i]-now);
            now*=c;
            if(now>=1e15)
            {
    
    
                now=-1;
                break;
            }
        }
        if(now==-1)
        {
    
    
             break;
        }
        else
        {
    
    
            res=min(res,sum);
        }
    }
    cout<<res<<endl;
}

非常重要的一个地方就是，now可以放在循环里，不用在浪费复杂度单独计算了