Codeforces 467C. George and Job, codeforces467c

Codeforces 467C. George and Job, codeforces467c

DP ....

C. George and Jobtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard output

The new ITone 6 has been released recently and George got really keen to buy it. unfortunately, he didn't have enough money, so George was going to work as a programmer. now he faced the following problem at the work.

Given a sequenceNIntegersP1, bytes,P2, middle..., middle ,...,PN. You are to chooseKPairs of integers:

[ L1, bytes, R1], region [ L2, bytes, R2], numbers..., numbers [ L K, Bytes, R K] (1 limit ≤ limit L1 bytes ≤ bytes R1 worker <worker L2 bytes ≤ bytes R2 rows <values... rows <values L KLimit ≤ limit R KLimit ≤ limit N; R IAccept-Encoding- L IBytes + bytes 1 records = bytes M), Then ),

In such a way that the value of sum is maximal possible. Help George to interact with the task.

Input

The first line contains three integersN,MAndK(1 limit ≤ limit (MLimit × limitK) Bytes ≤ bytesNLimit ≤00005000). The second line containsNIntegersP1, bytes,P2, middle..., middle ,...,PN(0 bytes ≤ bytesPILimit ≤ limit 109 ).

Output

Print an integer in a single line-the maximum possible value of sum.

Sample test (s) input

5 2 11 2 3 4 5

Output

9

Input

7 1 32 10 7 18 5 33 0

Output

61

/** * Created by ckboss on 14-9-19. */import java.util.*;public class GeorgeandJob {    static int n,m,k;    static long[] a = new long[5050];    static long[] sum = new long[5050];    static long[][] dp = new long[5050][3];    public static void main(String[] args){        Scanner in = new Scanner(System.in);        n=in.nextInt(); m=in.nextInt(); k=in.nextInt();        for(int i=1;i<=n;i++){            a[i]=in.nextInt();            sum[i]=sum[i-1]+a[i];        }        for(int i=m;i<=n;i++){            dp[i][1]=Math.max(dp[i-1][1],sum[i]-sum[i-m]);        }        for(int j=2;j<=k;j++){            for(int i=j*m;i<=n;i++){                dp[i][j%2]=Math.max(dp[i-m][(j-1)%2]+sum[i]-sum[i-m],dp[i-1][j%2]);            }        }        long ans=0;        for(int i=k*m;i<=n;i++){            ans=Math.max(ans,dp[i][k%2]);        }        System.out.println(ans);    }}