Best Cow Fences

题目描述

Farmer John's farm consists of a long row of N (1 <= N <= 100,000)fields. Each field contains a certain number of cows, 1 <= ncows <= 2000. 
FJ wants to build a fence around a contiguous group of these fields in order to maximize the average number of cows per field within that block. The block must contain at least F (1 <= F <= N) fields, where F given as input. 
Calculate the fence placement that maximizes the average, given the constraint. 

输入

* Line 1: Two space-separated integers, N and F. 
* Lines 2..N+1: Each line contains a single integer, the number of cows in a field. Line 2 gives the number of cows in field 1,line 3 gives the number in field 2, and so on. 

输出

* Line 1: A single integer that is 1000 times the maximal average.Do not perform rounding, just print the integer that is 1000*ncows/nfields. 

样例输入

10 6
6
4
2
10
3
8
5
9
4
1

样例输出

6500

分析：据说可以斜率DP，看别人题解的时候觉得很有道理，但是就是WA可还行。后来被逼无奈二分答案+DP了。

#include <iostream>
#include <string>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <deque>
#include <map>
#define range(i,a,b) for(int i=a;i<=b;++i)
#define LL long long
#define rerange(i,a,b) for(int i=a;i>=b;--i)
#define fill(arr,tmp) memset(arr,tmp,sizeof(arr))
using namespace std;
int n,f;
double ss[],aa[],Left=0x7fffffff,Right;
void init(){
    cin>>n>>f;
    range(i,,n){
        cin>>aa[i];
        ss[i]+=ss[i-]+aa[i];
        Left=min(Left,aa[i]);
        Right=max(Right,aa[i]);
    }
}
bool judge(double val){
    double tmp,pre=ss[f-]-(f-)*val;
    range(i,f,n){
        tmp=(ss[i]-ss[i-f])-f*val;
        pre+=aa[i]-val;
        pre=max(tmp,pre);
        if(pre>-1e-)return true;
    }
    return false;
}
void solve(){
    while(Right-Left>1e-){
        double mid=(Right+Left)/;
        if(judge(mid))Left=mid;
        else Right=mid;
    }
    cout<<(int)(Right*)<<endl;
}
int main() {
    init();
    solve();
    return ;
}