Codeforces 584 D. Dima and Lisa ( Codeforces Round #324 (Div. 2))

D. Dima and Lisa

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Dima loves representing an odd number as the sum of multiple primes, and Lisa loves it when there are at most three primes. Help them to represent the given number as the sum of at most than three primes.

More formally, you are given an odd numer n. Find a set of numbers pi (1 ≤ i ≤ k), such that

1. 1 ≤ k ≤ 3
2. pi is a prime
3. 

The numbers pi do not necessarily have to be distinct. It is guaranteed that at least one possible solution exists.

Input

The single line contains an odd number n (3 ≤ n < 109).

Output

In the first line print k (1 ≤ k ≤ 3), showing how many numbers are in the representation you found.

In the second line print numbers pi in any order. If there are multiple possible solutions, you can print any of them.

Sample test(s)

input

27

output

35 11 11

Note

A prime is an integer strictly larger than one that is divisible only by one and by itself.

题目大意：

就是给定一个数 m ，让你将其化为 <=3 个数的素数之和，

然后输出几个数 k， 和相应的素数

解题思路：

根据歌德巴赫猜想，可以推断出

任一大于2的偶数都可写成两个质数之和。

任一大于7的奇数都可写成三个素数之和。

然后再进行一下剪枝，

1）如果 m 是素数，那么输出 1  m

2）如果 m-2 是素数，那么输出的是2   2   m-2 

3）否则的话就是两个循环搞定（其实当时我以为是TLE的，但是竟然没有，嘿嘿~~）

上代码:

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <vector>#include <queue>#include <algorithm>#include <set>using namespace std;#define MM(a) memset(a,0,sizeof(a))typedef long long LL;typedef unsigned long long ULL;const int maxn = 50+5;const int mod = 1000000007;const double eps = 1e-7;bool isprime(int x){    if(x == 1)        return false;    for(int i=2; i*i<=x; i++)        if(x%i == 0)        return false;    return true;}/**任一大于2的偶数都可写成两个质数之和。任一大于7的奇数都可写成三个素数之和。**/int main(){    int m;    cin>>m;    if(isprime(m))        cout<<1<<endl<<m<<endl;    else if(isprime(m-2))    cout<<2<<endl<<2<<" "<<m-2<<endl;    else    {        puts("3");        bool ok = false;        for(int i=3; i<=m; i+=2)        {            if(ok)                break;            for(int j=3; j<=m; j+=2)            {                if(isprime(i) && isprime(j) && isprime(m-i-j))                {                    ok = true;                    cout<<i<<" "<<j<<" "<<m-i-j<<endl;                    break;                }            }        }    }    return 0;}