CodeCraft-19 and Codeforces #537 (Div. 2) B. Average Superhero Gang Power(枚举,贪心)

题目:https://codeforces.com/problemset/problem/1111/B

题意 给n个数,每个数最多操作k次,一共可以操作m次 操作分两种 一个是删除一个数 ,一个是另一个数+1 让最后的平均值最大

排序下记录下当前n个数的和,然后枚举删除多少个数时值最大,贪心先删除小的数

#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<iostream>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<stdlib.h>
#include<algorithm>
#include<time.h>
#include<unordered_map>
#define bug1(g) cout<<"test: "<<g<<endl
#define bug2(g,i) cout<<"test: "<<g<<" "<<i<<endl
#define bug3(g,i,k) cout<<"test: "<<g<<" "<<i<<" "<<k<<endl
using namespace std;
typedef long long ll;
int n,k,m;
int a[100005];
int main()
{
   scanf("%d %d %d",&n,&k,&m);
   ll sum=0;
   double maxx=0;
   for(int i =0;i<n;i++)
       {
           scanf("%d",&a[i]);
           sum+=a[i];
       }
   sort(a,a+n);
   for(int i =0;i<n;i++)
   {
       ll t=min((ll)m,1ll*k*(n-i)); //剩下的n-i个数最多还能增加多少
       double ans = (t+sum*1.0)/((n-i)*1.0);
       maxx=max(ans,maxx);
       if(m==0) break;
       m--;
       sum-=a[i];
   }
   printf("%.15f\n",maxx);
   return 0;
}