codeforces 578 C. Weakness and Poorness(三分)

題目鏈接：https://codeforc.es/contest/578/problem/C

思路：很明顯滿足三分要求，直接三分就行

#include <cstdio>
#include <cstdlib>
#include <cassert>
#include <cstring>
#include <bitset>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef vector<long long> VI;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define SZ(x) ((long long)(x).size())
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%lld",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define mod ll(1e9+7)
#define pb push_back
#define eps 1e-12
#define lc d<<1
#define rc d<<1|1
#define Pll pair<ll,ll>
#define P pair<int,int>
#define pi acos(-1)
int n;
double a[200008],b[200008],l,r,m1,m2;
double f(double x)
{
    double ma=-inff,q=0,w=0;
    FOR(i,1,n)
    {
        b[i]=a[i]-x,b[i]+=b[i-1];
        ma=max(ma,fabs(b[i]-q));ma=max(ma,fabs(b[i]-w));
        q=min(q,b[i]),w=max(w,b[i]);
    }
    return ma;
}
int main()
{
    cin.tie(0);
    cout.tie(0);
    cin>>n;
    FOR(i,1,n) sd(a[i]);
    l=-1000000.0,r=1000000.0;
    FOR(i,1,100)
    {
        m1=l+(r-l)/3.0;m2=r-(r-l)/3.0;
        if(f(m2)-f(m1)<=eps) l=m1;
        else r=m2;
    }
    printf("%.10lf\n",f(l));
    return 0;
}