Sleeping

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 1820 Accepted Submission(s): 671

Problem Description

ZZZ is an enthusiastic ACMer and he spends lots of time on training. He always stays up late for training. He needs enough time to sleep, and hates skipping classes. So he always sleeps in the class. With the final exams coming, he has to spare some time to listen to the teacher. Today, he hears that the teacher will have a revision class. The class is N (1 <= N <= 1000) minutes long. If ZZZ listens to the teacher in the i-th minute, he can get Ai points (1<=Ai<=1000). If he starts listening, he will listen to the teacher at least L (1 <= L <= N) minutes consecutively. It`s the most important that he must have at least M (1 <= M <= N) minutes for sleeping (the M minutes needn`t be consecutive). Suppose ZZZ knows the points he can get in every minute. Now help ZZZ to compute the maximal points he can get.

Input

The input contains several cases. The first line of each case contains three integers N, M, L mentioned in the description. The second line follows N integers separated by spaces. The i-th integer Ai means there are Ai points in the i-th minute.

Output

For each test case, output an integer, indicating the maximal points ZZZ can get.

Sample Input

10 3 3
1 2 3 4 5 6 7 8 9 10

Sample Output

49

Source

2011 Multi-University Training Contest 7 - Host by ECNU

璇渣開始補dp Q_Q

n分鐘的課聽第i分鐘可獲得ai分睡覺不得分

至少睡m分，每次聽課必須連續>=L 分鐘，求最大得分。

dp[i][j][0] 表示前 i 分鐘睡覺了 j 分鐘而且第 i 分鐘沒有睡覺，dp[i][j][1] 表示前 i 分鐘睡覺了 j 分鐘而且第 i 分鐘在睡覺。則：
dp[i][j][1] = max(dp[i-1][j-1][0], dp[i-1][j-1][1]);
dp[i][j][0] = max(dp[i-L][j][1]+sum[i]-sum[i-L], dp[i-1][j][0]+a[i]);

/** Author: ☆·aosaki(*’(OO)’*)  niconiconi★ **/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
//#include<bits/stdc++.h>
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <functional>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <list>
#include <stack>
//#include <tuple>
#define mem(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define lp(k,a) for(int k=1;k<=a;k++)
#define lp0(k,a) for(int k=0;k<a;k++)
#define lpn(k,n,a) for(int k=n;k<=a;k++)
#define lpd(k,n,a) for(int k=n;k>=a;k--)
#define sc(a) scanf("%d",&a)
#define sc2(a,b) scanf("%d %d",&a,&b)
#define lowbit(x) (x&(-x))
#define ll long long
#define pi pair<int,int>
#define vi vector<int>
#define PI acos(-1.0)
#define pb(a) push_back(a)
#define mp(a,b) make_pair(a,b)
#define TT cout<<"*****"<<endl;
#define TTT cout<<"********"<<endl;
inline int gcd(int a,int b)
{
    return a==0?b:gcd(b%a,a);
}

#define INF 1e9
#define eps 1e-8
#define mod 10007
#define MAX 1002

using namespace std;


int dp[1005][1005][2];
int a[1005],sum[1005];

int main()
{
    //freopen("in.txt","r",stdin);
    int n,m,L;
    while (~scanf("%d%d%d",&n,&m,&L))
    {
        mem(sum); mem(dp);
        lp(i,n)
          sc(a[i]);
        lp(i,n)
            sum[i]=sum[i-1]+a[i];
        lp(i,n)
        {
            if(i>=L)
                dp[i][0][0]=sum[i];
            for (int j=1;j<=m && j<=i;j++)
            {
                dp[i][j][1]=max(dp[i-1][j-1][0], dp[i-1][j-1][1]);
                if(i-L>=j)
                  dp[i][j][0]=max(dp[i-L][j][1]+sum[i]-sum[i-L], dp[i-1][j][0]+a[i]);
            }
        }
        printf("%d\n",max(dp[n][m][0],dp[n][m][1]));
    }
}