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 = 100005;
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 = 1; i <= n; ++i){
            scanf("%I64d",sum + i);
            sum[i] += sum[i-1];
        }
        double ret = 0;
        hd = tl = 0;
        for(int i = k; i <= n; ++i){
            while(hd + 1 < tl && check(q[tl-2],q[tl-1],i - k)) --tl;
            q[tl++] = i - k;
            while(hd + 1 < tl && check(q[hd+1],q[hd],i)) ++hd;
            ret = max(ret,(sum[i] - sum[q[hd]])*1.0/(i - q[hd]));
        }
        printf("%.2f
",ret);
    }
    return 0;
}