cf#320 Div.2 Problem E Weakness and Poorness

这题的代码是参考比赛中大神的AC代码的，据说用到了三分思想和前缀和，这里先记下来，日后再消化

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
int N;
int A[200005];
double f(double x){
    double res=0,s1=0,s2=0;
    for(int i=0;i<N;i++){
        if(s1<0)s1=A[i]-x;
        else s1+=A[i]-x;
        if(s2>0)s2=A[i]-x;
        else s2+=A[i]-x;
        res=max(res,max(fabs(s1),fabs(s2)));
    }
    return res;
}
int main(){
    while(scanf("%d",&N)!=EOF){
        for(int i=0;i<N;i++) scanf("%d",&A[i]);
        double l=-10000,r=10000,m1,m2,a1,a2,ans=1e233;
        for(int i=0;i<128;i++){
            m1=l+(r-l)/3;
            m2=l+(r-l)/3*2;
            a1=f(m1),a2=f(m2);
            ans=min(ans,min(a1,a2));
            if(a1<a2)r=m2;
            else l=m1;
        }
        printf("%.15lf",ans);
    }
    return 0;
}