Codeforces Round #666 (Div. 2)B. Power Sequence【暴力】

题目

传送门
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.

输入
3
1 3 2
输出
1
输入
3
1000000000 1000000000 1000000000
输出
1999982505

题意：给你一个序列，你可以对序列任意排列，然后你可以对序列中的任意元素进行一步操作：对任意元素进行加一，减一.问：你想让序列a0~an-1分别对应一个整数C的0-（n-1）次方，需要的最小操作数是多少？

思路：我们把序列从小到大排序，当n为1时，那个数一定对应的C的0次方，此时答案为a[0]-1，n=2时，我们选c=a[1],那么操作数还是a[0]-1,对于更大的数我们可以直接暴力反正范围只有1e9，2的32次方就可以，C更大时次数更少，具体看代码

AC code

#include<iostream>
#include<algorithm>
#include<cstring>
#include<map>
#include<cmath>
#include<queue>
#include<sstream>
#include<vector>
using namespace std;
#define ll long long
const ll inf=1e18;
ll a[120000],n;
void solve()
{
    ll s=0,ss=0;
    cin>>n;
     for(int i=0;i<n;i++)
       {cin>>a[i];
       ss+=a[i];}
       sort(a,a+n);
       if(n>=53)
        printf("%lld\n",ss-n);
       else
       {
           if(n==1||n==2)
              printf("%d\n",a[0]-1);
           else
           {
           ll maxx=inf,flag=1,oo=0;
          for(ll i=2;i<=100000;i++)
           {
               ll q=0,tt=1;
               for(ll j=0;j<n;j++)
               {

                  if(tt>1e14)
                  {
                      flag=0;
                      break;
                  }
                  q+=abs(a[j]-tt);
                  tt*=i;
               }
               if(flag==0)break;
              if(maxx>q)
              {
                  oo=i;
                  maxx=q;
              }
           }
           ll w=min(maxx,ss-n);
           printf("%lld\n",w);
           }
       }
}
int main()
{
  solve();////

}

不知道为什么，用pow就过不了下面贴上错误代码

#include<iostream>
#include<algorithm>
#include<cstring>
#include<map>
#include<cmath>
#include<queue>
#include<sstream>
#include<vector>
using namespace std;
#define ll long long
const ll inf=1e18;
int a[120000],n;
void solve()
{
    ll s=0,ss=0;
    cin>>n;
     for(int i=0;i<n;i++)
       {cin>>a[i];
       ss+=a[i];}
       sort(a,a+n);
       if(n>=53)
        printf("%lld\n",ss-n);
       else
       {
           if(n==1||n==2)
              printf("%d\n",a[0]-1);
           else
           {
           ll maxx=inf,flag=1,oo=0;
          for(ll i=2;i<=100000;i++)
           {
               ll q=0;
               for(ll j=0;j<n;j++)
               {
                  ll p=pow(i,j);
                  if(p>1e14)
                  {
                      flag=0;
                      break;
                  }
                  q+=abs(a[j]-p);
               }
               if(flag==0)break;
              if(maxx>q)
              {
                  oo=i;
                  maxx=q;
              }
           }
           ll w=min(maxx,ss-n);
           printf("%lld\n",w);
           }
       }
}
int main()
{
  solve();////
 
}