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

刺骨的言语ヽ痛彻心扉 2022-08-08 14:48 137阅读 0赞

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;
    }