Taxes CodeForces - 735D

Mr. Funt now lives in a country with a very specific tax laws. The total income of mr. Funt during this year is equal to
n (n ≥ 2) burles and the amount of tax he has to pay is calculated as the maximum divisor of
n (not equal to n, of course). For example, if
n = 6 then Funt has to pay
3 burles, while for n = 25 he needs to pay
5 and if n = 2 he pays only
1 burle.

As mr. Funt is a very opportunistic person he wants to cheat a bit. In particular, he wants to split the initial
n in several parts
n1 + n2 + ... + nk = n (here
k is arbitrary, even
k = 1 is allowed) and pay the taxes for each part separately. He can't make some part equal to
1 because it will reveal him. So, the condition
ni ≥ 2 should hold for all
i from 1 to
k.

Ostap Bender wonders, how many money Funt has to pay (i.e. minimal) if he chooses and optimal way to split
n in parts.

Input

The first line of the input contains a single integer n (2 ≤ n ≤ 2·109) — the total year income of mr. Funt.

Output

Print one integer — minimum possible number of burles that mr. Funt has to pay as a tax.

Example

Input

4

Output

2

Input

27

Output

3

题意：给出一个n，这个n可以分解成 n = n1 + n2 + …… + nk，其中k可以取任意数。要使得分解以后所有的n的最大因子（不包括自己本身）的和最小，问最小的和是多少。

思路：比赛的时候想到全部拆成素数是最好的，但是不知道怎么拆，看别人跑的特别快，就知道是数论题，绝望之下试了两发暴力，都是TLE了，GG。早上起来才知道有“哥德巴赫猜想”这个东西。

内容大概是如下两点：

1、所有大于2的偶数可以被分解成两个素数。

2、所有大于7的奇数可以被分解成三个素数。（n-3）为偶数，3是一个素数，所以是三个。

所以知道这个猜想之后就变得简单了：

1、偶数：n为2，答案是1，否则答案是2.

2、奇数：首先，n最少可以拆成三个素数，还有两种情况要考虑：n本身是一个素数的话答案就是1，n-2是一个素数答案就是2（一个奇数可以拆成一个偶数+一个奇数，偶数只有2是素数）。

#include <cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
int is_prime(int x){
for(int i=2;i<=(int)sqrt(x);i++){
if(x%i==0)
return 0;
}
return 1;
}

int main()
{
int n;
scanf("%d",&n);
int ans;
if(n==2||n==3||n==5||n==7)
ans=1;
else if(n%2==0)
ans=2;
else if(n&1){
if(is_prime(n))
ans=1;
else if(is_prime(n-2))
ans=2;
else
ans=3;
}
printf("%d\n",ans);
return 0;
}