【POJ 2018】 Best Cow Fences

【题目链接】

          http://poj.org/problem?id=2018

【算法】

            二分平均值

            检验时将每个数减去二分的值，求长度至少为L的子序列和的最大值，判断是否大于0

【代码】

          

#include <algorithm>  
#include <bitset>  
#include <cctype>  
#include <cerrno>  
#include <clocale>  
#include <cmath>  
#include <complex>  
#include <cstdio>  
#include <cstdlib>  
#include <cstring>  
#include <ctime>  
#include <deque>  
#include <exception>  
#include <fstream>  
#include <functional>  
#include <limits>  
#include <list>  
#include <map>  
#include <iomanip>  
#include <ios>  
#include <iosfwd>  
#include <iostream>  
#include <istream>  
#include <ostream>  
#include <queue>  
#include <set>  
#include <sstream>  
#include <stdexcept>  
#include <streambuf>  
#include <string>  
#include <utility>  
#include <vector>  
#include <cwchar>  
#include <cwctype>  
#include <stack>  
#include <limits.h> 
using namespace std;
#define MAXN 100010
const double eps = 1e-5;

int n,f,i;
double a[MAXN],b[MAXN],s[MAXN];
double l,r,mid,ans;

inline bool check(double x)
{
        int i;
        double mn = 1e6,ret = -1e6;
        for (i = 1; i <= n; i++) b[i] = a[i] - x;
        for (i = 1; i <= n; i++) s[i] = s[i-1] + b[i];
        for (i = f; i <= n; i++)
        {
                mn = min(mn,s[i-f]);
                ret = max(ret,s[i]-mn);
        }        
        return ret >= 0;
}

int main() {
        
        scanf("%d%d",&n,&f);
        for (i = 1; i <= n; i++) scanf("%lf",&a[i]);
        l = -1e6; r = 1e6;
        while (r - l > eps)
        {
                mid = (l + r) / 2;
                if (check(mid)) l = mid;
                else r = mid;
        }
        printf("%d
",int(r*1000));
        
        return 0;
    
}