2019 Shenyang F-Honk`s pool game network

F-Honk`s pool

Topic link: https: //nanti.jisuanke.com/t/41406

Title effect: honk pool has n, i-th tank has a [i ] liters per day will do the following operations honk

1. Take one liter of water up from the water pool (if there are a plurality of most pool water, an arbitrary choice) from 1L of

2. Find the least a water pool (if there are a plurality of minimum water pool, an arbitrary choice), the liters of water that was poured wherein

After you ask k days, the pool water is the number of poor

Problem-solving ideas:

1. simulated violence, as to why this can too, I do not know, certainly the cause of water data, look at the code

https://blog.csdn.net/weixin_44170305/article/details/100831508

2. dichotomy

During this operation, all of the average pool will be certainly moved, first requires the average value Ave, the n in the pond

The transfer amount of water is constant is k, we first separated the maximum amount of water is greater than two in the k ave days can be transferred, and then half of the pool is less than ave

Who can get the maximum amount you can subtract final

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6+10;
typedef long long ll;
ll a[maxn];
ll n,k;
bool check(ll mid,bool flag=false)
{
    ll sum=0;
    for(int i=1;i<=n;i++)
    {
        if(a[i]>=mid&&flag==false) sum+=a[i]-mid;
        else if(a[i]<=mid&&flag==true)   sum+=mid-a[i];
    }
    if(sum<=k) return true;
    else return false;
}
int main()
{
std::ios::sync_with_stdio(false);
while(std::cin>>n>>k)
{
ll top,low,sum=0;
for(int i=1;i<=n;i++) cin>>a[i],sum+=a[i];
sort(a+1,a+1+n);
ll maxx=a[n];
ll minn=a[1];
if(sum%n==0)
{
    top=low=sum/n;
}
else
{
    low=sum/n+1;
    top=sum/n;
}
ll ans=0;
ll ls=low,rs=maxx;
while(ls<=rs)
{
    ll mid=(ls+rs)>>1;
    if(check(mid))
    {
        ans=mid;
        rs=mid-1;
    }
    else ls=mid+1;
}
ll ans1=0;
ls=minn,rs=top;
while(ls<=rs)
{
    ll mid=(ls+rs)>>1;
    if(check(mid,true))
    {
        ans1=mid;
        ls=mid+1;
    }
    else rs=mid-1;
}
cout<<ans-ans1<<endl;
}
return 0;
}