Bachgold problem is very easy to formulate. Given a positive integer n represent it as a sum of maximum possible number of prime numbers. One can prove that such representation exists for any integer greater than 1.
Recall that integer k is called prime if it is greater than 1 and has exactly two positive integer divisors — 1 and k.
The only line of the input contains a single integer n (2 ≤ n ≤ 100 000).
The first line of the output contains a single integer k — maximum possible number of primes in representation.
The second line should contain k primes with their sum equal to n. You can print them in any order. If there are several optimal solution, print any of them.
5
2 2 3
6
3 2 2 2
题意:给一个数n,问最多由几个素数组成
解题思路:因为要尽可能多的素数,那么就尽量都是2
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
using namespace std;
#define LL long long
int n;
int main()
{
while(~scanf("%d",&n))
{
if(n==1) printf("0\n\n");
else if(n%2==1)
{
printf("%d\n",n/2);
while(n>3)
{
printf("2 ");
n-=2;
}
printf("3\n");
}
else
{
printf("%d\n2",n/2);
n-=2;
while(n>=2)
{
printf(" 2");
n-=2;
}
printf("\n");
}
}
return 0;
}