import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; import java.io.PrintWriter; import java.util.ArrayList; import java.util.Arrays; import java.util.Collections; import java.util.HashMap; import java.util.List; import java.util.Map; import java.util.StringTokenizer;

public class Main { public static void main(String[] args) throws IOException { FastScanner fs = new FastScanner(); PrintWriter out = new PrintWriter(System.out);

	int n = fs.nextInt();
	int k = fs.nextInt();
	int[] s = new int[n + 1];
	for(int i = 0;i < n;i++) {
		s[i + 1] = s[i] + fs.nextInt();
	}
	int maxNum = 0;
	for(int r = k + 1;r < n + 1;r++) {
		maxNum = Math.max(maxNum, s[r] - s[r - k - 1]);
	}
	out.println(maxNum);
	out.flush();
}

static class FastScanner {
	BufferedReader br;
	StringTokenizer st;

	public FastScanner() {
		br = new BufferedReader(new InputStreamReader(System.in));
	}

	boolean hasNext() throws IOException {
		while (st == null || !st.hasMoreTokens()) {
			String line = br.readLine();
			if (line == null)
				return false;
			st = new StringTokenizer(line);
		}
		return true;
	}

	public String next() throws IOException {
		hasNext();
		return st.nextToken();
	}

	public int nextInt() throws IOException {
		return Integer.parseInt(next());
	}

	public long nextLong() throws IOException {
		return Long.parseLong(next());
	}

	public double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}

	public String readLine() throws IOException {
		return br.readLine();
	}
}

}

#include<bits/stdc++.h>
using namespace std;
int a[1000005];
int sum[1000005];
int ans=0;


int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,k;
    cin>>n>>k;
    for(int i=1;i<=n;++i)
    {
        cin>>a[i];
        sum[i]=sum[i-1]+a[i];
    }
    for(int i=1;i<=n-k;++i)
    {
        ans=max(ans,(sum[i+k]-sum[i-1]));
    }
    cout<<ans;

}

一道很标准的一维前缀和题，由于倒水是只能往下一个杯子里倒，并且没有容量上限，所以问题可以简化为求k长区间和的最大值问题，只需先预处理前缀和数组，并枚举所有可能的左边界l，利用前缀和数组找出和最大的值输出即可。复杂度O(n)。

import sys
input = lambda:sys.stdin.readline().strip()

n=int(input())
k=int(input())
a=[0]+list(map(int,input().split()))
s=[0]*len(a)

for i in range(1,n+1):
    s[i]=s[i-1]+a[i]

ans=0
for l in range(1,n-k+1):
    r=l+k
    ans=max(ans,s[r]-s[l-1])
print(ans)

import sys data = sys.stdin.buffer.read().split() n = int(data[0]) k = int(data[1]) waters = list(map(int, data[2:2 + n])) len_waters = len(waters) pre_sum=[0]*(len_waters+1) for i in range(1,len_waters+1): pre_sum[i]=pre_sum[i-1]+waters[i-1] max_water=0 for p in range(1,len_waters+1): start=max(1,p-k) total = pre_sum[p] - pre_sum[start - 1] max_water = max(max_water, total) print(max_water)

import sys data = sys.stdin.buffer.read().split() n = int(data[0]) k = int(data[1]) waters = list(map(int, data[2:2 + n])) window_sum=0 len_waters = len(waters) max_waters = 0 left=0 right=0 while right<len_waters: c=waters[right] window_sum+=c right+=1 while right-left>k+1: window_sum-=waters[left] left+=1 max_waters=max(max_waters,window_sum) print(max_waters)