数论
拆成素数的话，答案就是1
根据哥德巴赫猜想，大于2的偶数都可以变成两个素数的和。
大于5d奇数，都可以拆成三个素数的和。
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
#include<bits/stdc++.h> using namespace std; long long n; bool check(int x){ for(int i=2;i*i<=x;i++) { if(x%i==0)return false; } return true; } int main() { scanf("%lld",&n); if(n>2&&n%2==0) { printf("2\n"); }else if(n==2) { printf("1\n"); } else { if(check(n)) cout<<"1"<<endl; else if(check(n-2)) cout<<"2"<<endl; else cout<<"3"<<endl; } }