BNUOJ 3958 MAX Average Problem

MAX Average Problem

Time Limit: 3000ms

Memory Limit: 65536KB

64-bit integer IO format: %lld      Java class name: Main

 

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<=M<=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

Source

2009 Multi-University Training Contest 19 - Host by BNU

Author

HK@Sphinx

 

解题：单调队列+斜率优化

 

 #include <stdio.h>
 #include <iostream>
 using namespace std;
 typedef long long LL;
 const int maxn = ;
 int q[maxn],hd,tl,n,k;
 LL sum[maxn];
 bool check(LL a,LL b,LL c){
     return (sum[c] - sum[b])*(b - a) <= (sum[b] - sum[a])*(c - b);
 }
 int main(){
     while(~scanf("%d%d",&n,&k)){
         for(int i = ; i <= n; ++i){
             scanf("%I64d",sum + i);
             sum[i] += sum[i-];
         }
         double ret = ;
         hd = tl = ;
         for(int i = k; i <= n; ++i){
             while(hd +  < tl && check(q[tl-],q[tl-],i - k)) --tl;
             q[tl++] = i - k;
             while(hd +  < tl && check(q[hd+],q[hd],i)) ++hd;
             ret = max(ret,(sum[i] - sum[q[hd]])*1.0/(i - q[hd]));
         }
         printf("%.2f\n",ret);
     }
     return ;
 }