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 ≤ k ≤ 3
- pi is a prime
The numbers pi do not necessarily have to be distinct. It is guaranteed that at least one possible solution exists.
The single line contains an odd number n (3 ≤ n < 109).
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.
27
3 5 11 11
A prime is an integer strictly larger than one that is divisible only by one and by itself.
任一大于7的奇数都可写成三个素数之和。
#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;
}
