【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-;

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 = ; i <= n; i++) b[i] = a[i] - x;
        for (i = ; i <= n; i++) s[i] = s[i-] + b[i];
        for (i = f; i <= n; i++)
        {
                mn = min(mn,s[i-f]);
                ret = max(ret,s[i]-mn);
        }
        return ret >= ;
}

int main() {

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

        return ;

}