Codeforces Round #382 (Div. 2) D. Taxes（分拆素数和）

题目链接：http://codeforces.com/contest/735/problem/D

D. Taxes
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.

Examples
input
4
output
2
input
27
output
3

【中文题意】
给出一个整数n，n可以由很多其他数的和组成，但不能出现1，当然也可以不拆分，然后问你：给你个n然后让你求n的最大因子为多少（不包括n本身），然后如果n被拆分的话，就是求拆分出来的这些数的因子和，规矩同上。
【思路分析】
哥德巴赫猜想：任何一个大于二的偶数都可以分解为两个素数和。
然后假如是奇数的话，直接特判下就好了。
【AC代码】

#include <cstdio>#include <algorithm>#include <cmath>#include <cstring>using namespace std;#define LL long longbool isprime(LL n){    for(LL i=2; i*i<=n; i++)    {        if((n%i)==0)        {            return false;        }    }    return true;}int main(){    LL n;    while(~scanf("%I64d",&n))    {        if(n==2)        {            printf("1\n");        }        else if(~n&1)        {            printf("2\n");        }        else        {            if(isprime(n)) printf("1\n");            else if(isprime(n-2))            {                printf("2\n");            }            else            {                printf("3\n");            }        }    }    return 0;}