动态规划：求连续的元素的和（素数个)

//题意：

 Rhezo and Prime Problems Attempted by:  3257 / Accuracy:  77% / Maximum Score:  20 /   630 Votes Tag(s):  

Data Structures, Dynamic Programming, Easy, Number Theory

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Rhezo and his friend Vanya love problem solving. They have a problem set containing $N$ problems, with points assigned to each. Rhezo wants to solve problems in such a way that he gets the maximum number of points. Rhezo has a weird habit of solving only prime number of consecutive problems, that is, if he solves $X$ consecutive problems from the problem set, then $X$ should be prime. Vanya has already solved all problems from the problem set and he wonders how much maximum points Rhezo can get. Vanya can answer this, but can you?

Input:

First line contains a single integer $N$, denoting the number of problems in the problem set. Next line contains $N$ space separated integers denoting the points assigned to the problems.

Output:

Print a single integer, the maximum points Rhezo can get.

Constraints:

$1\le N\le 5000$

$1\le$Points of Problems$\le {10}^5$

SAMPLE INPUT   4 8 1 3 7 SAMPLE OUTPUT   12

//代码：

import java.util.*; import java.io.*; import java.math.*; class Main { public static void main(String args[]) { //File file= new File("COMPLETE_FILE_PATH"); //InputStream is = new FileInputStream(file); //InputReader in = new InputReader(is); //InputReader in = new InputReader(System.in); //PrintWriter out = new PrintWriter(System.out); int t = 1; while (t-- > 0) { Task solver = new Task(); solver.solve(t); } } static class Task{ long mod=1000000007; static boolean[]isPrime; static List<Integer>Prime; public static void solve(int testNumber) { Scanner sc=new Scanner(System.in); int n=sc.nextInt(); long []arr=new long[n+1]; long []dp=new long[n+1]; for(int i=1;i<=n;i++) { arr[i]=arr[i-1]+sc.nextInt(); } Prime=new ArrayList<>(); judge(n); dp[0]=dp[1]=0; /* for(int i=2;i<=n;i++) dp[i]=dp[i-1];*/ for(int i=2;i<=n;i++) { dp[i]=dp[i-1]; for(int j=0;j<Prime.size()&&Prime.get(j)<=i;j++){ if(Prime.get(j)==i){ dp[i]=Math.max(dp[i],arr[i]); } else{ int p=i-Prime.get(j)-1; dp[i]=Math.max(dp[i],dp[p]+arr[i]-arr[p+1]); } } } System.out.println(dp[n]); } public static void judge(int N) { isPrime=new boolean[N+1]; for(int i=2;i<=N;i++){ isPrime[i]=true; } for(int i=2;i<=N;i++){ if(isPrime[i]){ Prime.add(i); } for(int j=i*i;j<=N;j+=i){ isPrime[j]=false; } } } } }https://doigetitnow.quora.com/Rhezo-and-Prime-problems