codeforces 578c//Weakness and Poorness// Codeforces Round #320 (Div. 1)

题意：一个数组arr,一个数字x,要使arr-x的最大子段最小，问该最小值。

三分x,复杂度logn,内层是最大子段的模板，只能用n复杂度的。因为是绝对值最大，正负各求一次，取大的。精度卡得不得了，要1e-12左右才能过。看着数据才调出精度的。

乱码：

//#pragma comment(linker,"/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
#include <stack>
#include <iomanip>
using namespace std;
const int SZ=,INF=0x7FFFFFFF;
const double EPS=8e-;
typedef long long lon;
double arr[SZ],b[SZ],dp[SZ];

double max(double a[],int n)
{
    double sum,maxsum;
    int i ;
    sum = maxsum = ;
    for(i = ;i<n;i++)
    {
        sum +=a[i];
        if(sum>maxsum)
            maxsum = sum;
        else if(sum<)
            sum = ;
    }
    return maxsum;
}

double work(int n)
{
    double res1=max(b,n);
    for(int i=;i<n;++i)b[i]=-b[i];
    double res2=max(b,n);
    return max(res1,res2);
}

int main()
{
    std::ios::sync_with_stdio();
    //freopen("d:\\1.txt","r",stdin);
    //for(;scanf("%d",&n)!=EOF;)
    {
        int n;
        clock_t bg=clock(),end;
        cin>>n;
        for(int i=;i<n;++i)
        {
            cin>>arr[i];
        }
        double lo=-,hi=,res=;

        for(;lo<hi&&fabs(lo-hi)>EPS;)
        {
            //end=clock();
            //if(end-bg>1500)break;
            double mid1=lo+(hi-lo)/;
            double mid2=lo+*(hi-lo)/;
            for(int i=;i<n;++i)
            {
                b[i]=arr[i]-mid1;
            }
            double res1=work(n);
            for(int i=;i<n;++i)
            {
                b[i]=arr[i]-mid2;
            }
            double res2=work(n);
            res=res1;
            //cout<<lo<<" "<<hi<<' '<<res<<endl;
            if(res1<res2)hi=mid2;
            else lo=mid1+1e-;
        }
        cout<<fixed<<setprecision()<<res<<endl;
    }
    return ;
}