P5242 [USACO19FEB]Cow Dating

题目链接

题意分析

首先我们可以得出计算公式

\[s_i=\prod_{k=1}^i(1-p_k)
\]

\[f_i=\sum_{k=1}^i\frac{p_k}{1-p_k}
\]

那么

\[ans(i,j)=\frac{s_r}{s_{l-1}}{f_r-f_{l-1}}
\]

强行枚举 \(O(n^2)\)

我们冷机观察一波发现 可以使用尺取法

然后优化成了\(O(n)\)

CODE:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdlib>
#include<string>
#include<queue>
#include<map>
#include<stack>
#include<list>
#include<set>
#include<deque>
#include<vector>
#include<ctime>
#define ll long long
#define inf 0x7fffffff
#define N 500008
#define IL inline
#define M 1008611
#define D long double
#define R register
using namespace std;
template<typename T>IL void read(T &_)
{
    T __=0,___=1;char ____=getchar();
    while(!isdigit(____)) {if(____=='-') ___=0;____=getchar();}
    while(isdigit(____)) {__=(__<<1)+(__<<3)+____-'0';____=getchar();}
    _=___ ? __:-__;
}
/*-------------OI使我快乐-------------*/
int n;
D num[M],cdy=1.0,wzy,ans;
int main()
{
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
    read(n);
    for(R int i=1,x;i<=n;++i)
    {
        read(x);
        num[i]=((D)x/1000000.0);
        ans=max(ans,num[i]);
    }
    for(R int i=1,tail=1;i<=n;++i)
    {
        while(tail<=n&&cdy*wzy<cdy*(1.0-num[tail])*(wzy+num[tail]/(1.0-num[tail])))
        {
            cdy*=(1.0-num[tail]);
            wzy+=num[tail]/(1.0-num[tail]);
            ++tail;
        }
        ans=max(ans,cdy*wzy);
        cdy/=(1.0-num[i]);wzy-=num[i]/(1.0-num[i]);
    }
    printf("%d\n",(int)(ans*1000000));
//	fclose(stdin);
//	fclose(stdout);
    return 0;
}

HEOI 2019 RP++