【链接】 我是链接,点我呀:)
【题意】
【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());
}
}
}