数据结构：HDU 2993 MAX Average Problem

MAX Average Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7639    Accepted Submission(s): 1667


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 6 6 4 2 10 3 8 5 9 4 1
 

 

Sample Output
6.50
 

　　这题有个模型，挺经典的。

　　然而，不知为何，HDU上这道题已经不能AC了，原来的标程都会TLE！！！

　　所以我就把我的TLE的程序当标程挂在这了~~~

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 using namespace std;
 7 int sum[1000010],q[1000010];
 8 int n,L;
 9 double mid;
10 double C(int i)
11 {
12     return sum[i]*1.0-mid*i;
13 }
14 int h,t;
15 bool IS_OK(double x)
16 {
17     h=1;t=0;
18     for(int i=L;i<=n;i++)
19     {
20         while(h<=t&&C(q[t])<=C(i))t--;
21         q[++t]=i;
22     }
23     for(int i=1;i<=n-L+1;i++)
24     {
25         if(q[h]-i+1<L)h++;
26         if(C(q[h])-C(i-1)>=0)
27             return true;
28     }
29     return false;
30 }
31 int main()
32 {
33     double lo,hi;
34     while(~scanf("%d%d",&n,&L)){
35         lo=1.0,hi=2000.0;
36         for(int i=1;i<=n;i++){
37             scanf("%d",&sum[i]);
38             lo=min(sum[i]*1.0,lo);
39             hi=max(sum[i]*1.0,hi);
40         }
41             
42         for(int i=2;i<=n;i++)
43             sum[i]+=sum[i-1];
44 
45         
46         while(hi-lo>=0.01)
47         {
48             mid=(lo+hi)/2.0;
49             if(IS_OK(mid))lo=mid;
50             else hi=mid;
51         }
52         printf("%.2f\n",hi);
53     }
54     return 0;
55 }

 

posted @ 2016-03-04 18:50  TenderRun  阅读(202)  评论(0)  编辑  收藏  举报