【Study Notes】CF1817E Half-sum

First, the optimal greedy strategy is: after sorting the original array , select a dividing point $p$ , then for$[1,p]$ merge from back to front,$[p+1,n]$ Merge from front to back, and take the difference between the two numbers before and after as the answer. Simulating addition in binary can do$O\left(n\right)$ checks. Note that there is no need to maintain high-precision addition evenly (doing so will result in one more$log)$ , because we don't care about the intermediate results.

Then, derive it from the perspective of a primary composite function (this form is more concise):

设$f(a_1,...,a_i,x)=ax+b$，那么$f(a_1,...,a_i,a_{i+1},x)=a\cdot \frac{a_{i+1}+x}{2}+b=\frac{1}{2}ax+\frac{1}{2}a\cdot a_{i+1}+b$

Note that $a=\frac{1}{2^i}$, so from the perspective of growth rate, only the top $30$ items and$30$ items are checked.

Usually, we can eliminate some solutions that are definitely not optimal through neighbor comparison (finding the variation ).

发现$f(a_1,...,a_{i+1})-f(a_1,...,a_i)=\frac{1}{2}a\cdot (a_{i+1}-a_i)=\frac{1}{2^i}(a_{i+1}-a_i)$ , so consider the difference array of the original array$d_i=a_{i+1}-a_i$, then from scheme $i$ tojjThe change of$j$$$-\frac{d_i}{2^i}-\sum _{i<k<j}{d}_k(\frac{1}{2^k}-\frac{1}{2^{n-k}})+\frac{d_j}{2^{n-j}}$$

Note that if $d_i=0nana$ ii_$i$ will definitely not become the optimal dividing point, so weshould satisfy$d_i>0$ . Then, when$i\le j\le \frac{n}{2}$When , the middle formula must be $\le 0$ , only when$n-j\le i+30$ years later may be the answer. That is:$j\ge n-i-30$ . In this way, we only need to find the first one that satisfies$\le \frac{n}{2}-30$ and$d_i>0$ position, this must be optimal; if it does not exist, then test$[\frac{n}{2}-30,\frac{n}{2}]$ in all locations. The equivalent implementation is, before checking$30$ items satisfy$d_i>0$ position. For$j>\frac{n}{2}$The situation is obviously the same.

Complexity $O(n\log V)$。

#include<bits/stdc++.h>
#define ll long long
#define pb push_back
#define fi first
#define se second
#define db double
#define inf 0x3f3f3f3f3f3f3f3f
using namespace std;
const int mod=1e9+7;
const int N=1e6+65;
int T,n,m,res;
ll a[N],inv2=(mod+1)/2;
ll b[N],c[N],r[N];
void solve(int x){
    
    
    for(int i=0;i<=n+60;i++)b[i]=c[i]=0;
    for(int i=x;i>=1;i--){
    
    
        int j=i-(i==x);
        b[n-j]+=a[i];
    }for(int i=0;i<=n+60;i++)b[i+1]+=b[i]/2,b[i]%=2;
    for(int i=x+1;i<=n;i++){
    
    
        int j=n-i+1-(i==x+1);
        c[n-j]+=a[i];
    }for(int i=0;i<=n+60;i++)c[i+1]+=c[i]/2,c[i]%=2;
    for(int i=0;i<=n+60;i++){
    
    
        c[i]-=b[i];if(c[i]<0)c[i]+=2,c[i+1]--;
    }
    int it=n+60;while(it&&c[it]==r[it])it--;
    if(c[it]>r[it]){
    
    
        for(int i=0;i<=n+60;i++)r[i]=c[i];
        res=x;
    }
}
int main(){
    
    
    ios::sync_with_stdio(false);
    cin.tie(0),cout.tie(0);
    cin>>T;
    while(T--){
    
    
        cin>>n;for(int i=1;i<=n;i++)cin>>a[i];
        sort(a+1,a+1+n);res=1;for(int i=0;i<=n+60;i++)r[i]=0;
        int j=1;
        for(int i=1;i<=30;i++){
    
    
            while(j<n&&a[j]==a[j+1])j++;
            if(j==n)break;solve(j),j++;
        }j=n;
        for(int i=1;i<=30;i++){
    
    
            while(j>1&&a[j]==a[j-1])j--;
            if(j==1)break;solve(j-1),j--;
        }ll x=a[res];
        for(int i=res-1;i>=1;i--)x=(x+a[i])*inv2%mod;
        ll y=a[res+1];
        for(int i=res+2;i<=n;i++)y=(y+a[i])*inv2%mod;
        cout<<(y-x+mod)%mod<<"\n";
    }
}