【Codeforces 584D】Dima and Lisa

【链接】 我是链接,点我呀:)
【题意】

让你把一个奇数n分成最多个质数的和

【题解】

10的9次方以内,任意两个质数之间的差距最大为300 因此可以这样,我们先从i=n-2开始一直递减直到i变成最大的p(p < n)且p是一个质数 这样的话我们就得到了一个质数p,和一个偶数n-p(因为p肯定是一个奇数) 然后n-p<=300,所以我们可以对于剩下的n-p进行暴力求解,分成两个数的和 会发现在小于等于300的时候,总是有成对的质数和为i的

【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 here
        new Task().solve(in, out);
        out.close();
    }
    
    static int N = 50000;
    static class Task{
        
    	boolean is(int x) {
    		int len = (int)Math.sqrt(x);
    		for (int i = 2;i <=len;i++)
    			if (x%i==0)
    				return false;
    		return true;
    	}
    	
        int n;
        public void solve(InputReader in,PrintWriter out) {
        	n = in.nextInt();
        	int len = (int)Math.sqrt(n);
        	if (is(n)) {
        		out.println(1);
        		out.println(n);
        		return;
        	}
        	//n不是质数,n是奇数
        	for (int i = n-2;i >= 1;i--) {
        		if (is(i)) {
        			int rest = n-i;
        			if (rest==2) {
        				out.println(2);
        				out.print(i+" "+rest);
        			}else {
        				for (int j = 2;j <= rest;j++) {
        					if (is(j)&&is(rest-j)) {
        						out.println(3);
        						out.print(i+" "+j+" "+(rest-j));
        						return;
        					}
        				}
        			}
        			break;
        		}
        	}
        }
    }

    

    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());
        }
    }
}