【链接】 我是链接,点我呀:)
【题意】

让你从1..n这n个数字中
选出来k个不相交的长度为m的区间
然后这个k个区间的和最大
求出这k个区间的和的最大值

【题解】

设dp[i][j]表示前i个数字已经选出了j个区间的最大值
看看是以当前位置为结尾选择一个区间,还是这个位置不包括在任何一个区间里.
分这两种情况转移就好

【代码】

import java.io.*;
import java.util.*;public class Main {static InputReader in;static PrintWriter out;public static void main(String[] args) throws IOException{//InputStream ins = new FileInputStream("E:\\rush.txt");InputStream ins = System.in;in = new InputReader(ins);out = new PrintWriter(System.out);//code start from herenew Task().solve(in, out);out.close();}static int N = 5000;static class Task{int n,m,k;int a[];long sum[];long dp[][];long get_sum(int l,int r) {return sum[r]-sum[l-1];}public void solve(InputReader in,PrintWriter out) {n = in.nextInt();m = in.nextInt();k = in.nextInt();a = new int[N+10];sum = new long[N+10];dp = new long[N+10][N+10];for (int i = 1;i <= n;i++) {a[i] = in.nextInt();}for (int i = 1;i <= n;i++) sum[i]=sum[i-1]+a[i];long ans = 0;dp[0][0] = 0;for (int i = 1;i <= n;i++)for (int j = 0;j <= Math.min(k, i);j++) {if (j==0) {dp[i][j] = Math.max(dp[i][j], dp[i-1][j]);}else {if (i-m>=0) {dp[i][j] = Math.max(dp[i][j], dp[i-m][j-1]+get_sum(i-m+1,i));dp[i][j] = Math.max(dp[i][j], dp[i-1][j]);}}ans = Math.max(ans, dp[i][j]);}out.println(ans);}}static class InputReader{public BufferedReader br;public StringTokenizer tokenizer;public InputReader(InputStream ins) {br = new BufferedReader(new InputStreamReader(ins));tokenizer = null;}public String next(){while (tokenizer==null || !tokenizer.hasMoreTokens()) {try {tokenizer = new StringTokenizer(br.readLine());}catch(IOException e) {throw new RuntimeException(e);}}return tokenizer.nextToken();}public int nextInt() {return Integer.parseInt(next());}}
}