Codeforces Round #320 (Div. 1) [Bayan Thanks-Round] C A Weakness and Poorness （三分）

显然f(x)是个凹函数，三分即可，计算方案的时候dp一下。eps取大了会挂精度，指定循环次数才是正解。

#include<bits/stdc++.h>

using namespace std;
const double eps = 1e-11;
const int maxn = 2e5+5;
double a[maxn];
double d1[maxn],d2[maxn];
int n;
inline double cal(double x)
{
    double a1 = 0., a2 = 0.;
    for(int i = 1; i <= n; i++){
        d1[i] = max(0.,d1[i-1])+a[i]-x;
        a1 = max(a1,d1[i]);
        d2[i] = min(0.,d2[i-1])+a[i]-x;
        a2 = min(a2,d2[i]);
    }
    return max(fabs(a1),fabs(a2));
}

int main()
{
    //freopen("in.txt","r",stdin);
    scanf("%d",&n);
    for(int i = 1; i <= n; i++) scanf("%lf",a+i);
    double L = -1e4, R = 1e4;
    double M1v,M2v;
    while(R-L>eps){
        double M1 = L+(R-L)/3, M2 = L+(R-L)/3*2;
        M1v = cal(M1), M2v = cal(M2);
        if(abs(M1v-M2v)<eps){
            L = M1; R = M2;
        }else {
            if(M1v > M2v){
                L = M1;
            }else {
                R = M2;
            }
        }
    }
    printf("%.15lf",(cal(L)+cal(R))/2);
    return 0;
}