hdoj MAX Average Problem 2993 （斜率优化DP）

MAX Average Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 7613    Accepted Submission(s): 1667


[align=left]Problem Description[/align]
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.
 

[align=left]Input[/align]
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].

 

[align=left]Output[/align]
For every test case, output one single line contains a real number, which is mentioned in the description, accurate to 0.01.
 

[align=left]Sample Input[/align]

10 6
6 4 2 10 3 8 5 9 4 1

 

[align=left]Sample Output[/align]

6.50//思路：http://www.docin.com/p-47950655.html一种新思想。。。//看网上的都是TLM。。。先贴上。。。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define INF 0x3f3f3f3f
#define ll long long
#define N 100010
using namespace std;
int sum
,s
,a
;
bool cross(int i,int j,int k)
{
if((sum[j]-sum[i])*(k-i)>=(sum[k]-sum[i])*(j-i))
return true;
return false;
}
double fx(int i,int t)
{
double tmp;
tmp=1.0*(sum[t]-sum[i])/(t-i);
return tmp;
}
int main()
{
int n,k;
int i;
while(scanf("%d%d",&n,&k)!=EOF)
{
sum[0]=0;
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
sum[i]=sum[i-1]+a[i];
}
int top=0,low=0;
double ans=0;
for(i=k;i<=n;i++)
{
int j=i-k;
while(top-low>=1&&cross(s[top-1],s[top],j))
top--;
s[++top]=j;
while(top-low>=1&&fx(s[low+1],i)>=fx(s[low],i))
low++;
ans=max(ans,fx(s[low],i));
}
printf("%.2lf\n",ans);
}
return 0;
}