Codeforces 467C. George and Job

最新推荐文章于 2020-01-23 19:29:44 发布

转载最新推荐文章于 2020-01-23 19:29:44 发布 · 51 阅读

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#java

本文介绍了一个编程问题，目标是在给定长度为n的整数序列中选择k个子序列（每个长度为m），使得这些子序列元素之和的最大值。通过动态规划的方法，实现了高效的求解算法。

DP....

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 sequence of n integers p1, p2, ..., pn. You are to choose k pairs of integers:

[l1, r1], [l2, r2], ..., [lk, rk] (1 ≤ l1 ≤ r1 < l2 ≤ r2 < ... < lk ≤ rk ≤ n; ri - li + 1 = m),

in such a way that the value of sum $\text{[math]}$ is maximal possible. Help George to cope with the task.

Input

The first line contains three integers n, m and k (1 ≤ (m × k) ≤ n ≤ 5000). The second line contains n integers p1, p2, ..., pn (0 ≤ pi ≤ 109).

Output

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

Sample test(s)

input
5 2 1
1 2 3 4 5

output
9

input
7 1 3
2 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);
    }
}