Codeforces Round #320 (Div. 1) C. Weakness and Poorness-蒲公英云

Codeforces Round #320 (Div. 1) C. Weakness and Poorness

具有很明显的单峰性质
直接三分，用动态规划的方式求出weekness
注意三分的时候，很多人用eps来判断三分结束，这样有一些精度误差
直接三分100次即可

// whn6325689
// Mr.Phoebe
// http://blog.csdn.net/u013007900
#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <climits>
#include <complex>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <bitset>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <ctime>
#include <set>
#include <map>
#include <cmath>
#include <functional>
#include <numeric>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
#define eps 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LLINF 1LL<<62
#define speed std::ios::sync_with_stdio(false);
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef complex<ld> point;
typedef pair<int, int> pii;
typedef pair<pii, int> piii;
typedef vector<int> vi;
#define CLR(x,y) memset(x,y,sizeof(x))
#define CPY(x,y) memcpy(x,y,sizeof(x))
#define clr(a,x,size) memset(a,x,sizeof(a[0])*(size))
#define cpy(a,x,size) memcpy(a,x,sizeof(a[0])*(size))
#define debug(a) cout << #a" = " << (a) << endl;
#define debugarry(a, n) for (int i = 0; i < (n); i++) { cout << #a"[" << i << "] = " << (a)[i] << endl; }
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define lowbit(x) (x&(-x))
#define MID(x,y) (x+((y-x)>>1))
#define ls (idx<<1)
#define rs (idx<<1|1)
#define lson ls,l,mid
#define rson rs,mid+1,r
template<class T>
inline bool read(T &n)
{
    T x = 0, tmp = 1;
    char c = getchar();
    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();
    if(c == EOF) return false;
    if(c == '-') c = getchar(), tmp = -1;
    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();
    n = x*tmp;
    return true;
}
template <class T>
inline void write(T n)
{
    if(n < 0)
    {
        putchar('-');
        n = -n;
    }
    int len = 0,data[20];
    while(n)
    {
        data[len++] = n%10;
        n /= 10;
    }
    if(!len) data[len++] = 0;
    while(len--) putchar(data[len]+48);
}
//-----------------------------------
const int MAXN=200010;
int A[MAXN],n;
double B[MAXN];
double solve(double x)
{
    for(int i=1; i<=n; i++)
        B[i]=A[i]*1.0-x;
    double ans=0,res=-INF;
    for(int i=1; i<=n; i++)
    {
        if(res<0)
            res=B[i];
        else
            res+=B[i];
        ans=max(ans,abs(res));
    }
    res=INF;
    for(int i=1; i<=n; i++)
    {
        if(res>0)
            res=B[i];
        else
            res+=B[i];
        ans=max(ans,abs(res));
    }
    return ans;
}
int main()
{
    while(read(n))
    {
        double l=INF,r=-INF;
        for(int i=1; i<=n; i++)
        {
            read(A[i]);
            l=min(l,A[i]*1.0);
            r=max(r,A[i]*1.0);
        }
        int cas=100;
        while(cas--)
        {
            double mid=(l+r)*0.5,midd=(mid+r)*0.5;
            if(solve(mid)<solve(midd))
                r=midd;
            else
                l=mid;
        }
        printf("%.10f\n",solve(l));
    }
    return 0;
}