做法

每次用最小的号来上分掉分即可
用优先队列或者multiset维护当前最小的号即可

代码


#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp(aa,bb) make_pair(aa,bb)
#define _for(i,b) for(int i=(0);i<(b);i++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,b,a) for(int i=(b);i>=(a);i--)
#define mst(abc,bca) memset(abc,bca,sizeof abc)
#define X first
#define Y second
#define lowbit(a) (a&(-a))
#define debug(a) cout<<#a<<":"<<a<<"\n"
typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
typedef long double ld;
const int N=100010;
const int INF=0x3f3f3f3f;
const int mod=1e9+7;
const double eps=1e-6;
const double PI=acos(-1.0);

int n,m;
multiset<ld> s;
ld sum;

void solve(){
    cin>>n>>m;
    rep(i,1,n){
        int x;cin>>x;
        s.insert(x);
        sum+=x;
    }
    while(m--){
        int x;cin>>x;
        ld t=*s.begin();
        if(x>=t){
            sum-=t;
            sum+=(x+t)/2;
            s.erase(s.begin());
            s.insert((x+t)/2);
            cout<<fixed<<setprecision(2)<<sum<<"\n";
            continue;
        }
        auto p=s.lower_bound(x);
        if(p==s.end()){
            t=*(--p);
            sum-=t;
            sum+=(x+t)/2;
            s.erase(p);
            s.insert((x+t)/2);
        }
        else{
            sum-=*p;
            sum+=(x+*p)/2;
            s.erase(p);
            s.insert((x+*p)/2);
        }
        cout<<fixed<<setprecision(2)<<sum<<"\n";
    }
}


int main(){
    ios::sync_with_stdio(0);cin.tie(0);
//    int t;cin>>t;while(t--)
    solve();
    return 0;
}