Codeforces Round #382 (Div. 2) 735D - Taxes

D. Taxes time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard 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 partsn1 + 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      

題意就是給你一個數N，然後把這個數分成任意份大小不同的數，把這個數看成一份也可以，還有就是每份都要大于1，你要怎麼分才能使每份數除自己外的最大因數加起來最小。要每個數除自己外的最大因數最小那就會想到素數，是以要盡量使這個給出的數分成幾份素數，這其中可以用到已被證明的哥德巴赫猜想：大于等于4的偶都可以由兩個相加得到。是以除2以外的偶數都隻需要輸出2即可，2的情況特判一下；如果是奇數，先判斷是否為素數，在判斷這個數N-2是否為素數，如果都不是的話我們還可用上面的哥德巴赫猜想推導出大于等于7的奇數都可以由3個素數相加得到，輸出3就行！

部落客就是在比賽時卡在了沒有判斷N-2為素數的情況，當時腦子秀逗了。代碼很短，就是腦洞不夠，好題贊一個！

代碼：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
bool judge(int key){
    for(int i = 2; i*i <= key; i ++){
        if(key%i == 0) return false;
    }
    return true;
}

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