D. Taxes
time limit per test
memory limit per test
input
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
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 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
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
歌德巴赫猜想:
1.任何一个大于等于6的偶数都可以分为两个奇质数之和
2.任何一个大于等于9的质数都可以分为三个奇质数之和
所以，当输入的一个数为质数时输出1，若为偶数输出2，若为奇数，则有两种形式，(1).奇数+奇数+奇数，(2)奇数+偶数，如果是第二种则把数减去2后判断是否为质数
#include <bits/stdc++.h>
#define maxn 200005
#define MOD 1000000007
using namespace std;
typedef long long ll;
bool judge(ll m){
for(ll i = 2; i * i <= m; i++){
if(m % i == 0)
return false;
}
return true;
}
int main(){
ll n;
scanf("%I64d", &n);
if(n == 2)
puts("1");
else if(n % 2 == 0)
puts("2");
else{
if(judge(n))
puts("1");
else{
ll m = n - 2;
if(judge(m))
puts("2");
else
puts("3");
}
}
return 0;
}
