HDU 2993

MAX Average Problem

Problem Description

Consider a simple sequence which only contains positive integers as a1, a2 ... an, and a number k. Define ave(i,j) as the average value of the sub sequence ai ... aj, i<=j. Let’s calculate max(ave(i,j)), 1<=i<=j-k+1<=n.

 

Input

There multiple test cases in the input, each test case contains two lines.
The first line has two integers, N and k (k<=N<=10^5).
The second line has N integers, a1, a2 ... an. All numbers are ranged in [1, 2000].

 

Output

For every test case, output one single line contains a real number, which is mentioned in the description, accurate to 0.01.

 

Sample Input

10 66 4 2 10 3 8 5 9 4 1

 

Sample Output

6.50

 

一道简单的斜率优化。用sum数组存放和，sum[i]为a[1]~a[i]的和,要求(sum[i]-sum[j])/(i-j)的最大值(i-j>=k)。

要用到输入挂，不然会超时。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;#define ll long longconst int maxn=100005;int sum[maxn];int q[maxn];namespace fastIO{#define BUF_SIZE 100000bool IOerror=0;inline char nc(){    static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;    if(p1==pend)    {        p1=buf;        pend=buf+fread(buf,1,BUF_SIZE,stdin);        if(pend==p1)        {            IOerror=1;            return -1;        }    }    return *p1++;}inline bool blank(char ch){    return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';}inline void read(int &x){    char ch;    while(blank(ch=nc()));    if(IOerror)        return ;    for(x=ch-'0'; (ch=nc())>='0'&&ch<='9'; x=x*10+ch-'0');}#undef BUF_SIZE}int main(){    int n,k;    while(fastIO::read(n),!fastIO::IOerror)        //while(~scanf("%d%d",&n,&k))    {        fastIO::read(k);        sum[0]=0;        for(int i=1; i<=n; i++)        {            int a;            fastIO::read(a);            //a = GetInt();            sum[i]=sum[i-1]+a;        }        int head=0,tail=0;        double ans=0;        for(int i=k; i<=n; i++)        {            int j=i-k;            while(head+1<tail)//维护下凸性            {                int x1=j-q[tail-1];                int x2=j-q[tail-2];                int y1=sum[j]-sum[q[tail-1]];                int y2=sum[j]-sum[q[tail-2]];                if((ll)x1*y2>=(ll)y1*x2)                    tail--;                else                    break;            }            q[tail++]=j;            while(head+1<tail)//维护上凸性            {                int x1=q[head+1]-q[head];                int x2=i-q[head+1];                int y1=sum[q[head+1]]-sum[q[head]];                int y2=sum[i]-sum[q[head+1]];                if((ll)x1*y2>=(ll)y1*x2)                    head++;                else                    break;            }            double tmp=1.0*(sum[i] - sum[q[head]]) / (i - q[head]);            ans=max(ans,tmp);        }        printf("%.2f\n",ans);    }    return 0;}