标签：答案 eal namespace opened contains sed input nts 分享
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.
The first line of the input contains a single integer n (2?≤?n?≤?2·109) — the total year income of mr. Funt.
Print one integer — minimum possible number of burles that mr. Funt has to pay as a tax.
读题易想到所有素数都只算1的税，所以联想到分解成素数（本身素数那就是1）。然后根据哥德巴赫猜想，所有偶数可以分解为两个素数的和，所以除了2以外的所有偶数答案都是2，而对于非素数奇数就是看他能不能分解2+素数的形式，否则就是3
#include <iostream> #include <iomanip> #include <cstring> #include <cstdio> #include <cmath> #include <algorithm> using namespace std; typedef long long LL; LL n; LL res = 0; bool isprime(LL n){ if (n == 2){ return true; } for (int i=2;i<=sqrt(n);i++){ if (n % i == 0){ return false; } } return true; } int main(){ // freopen("test.in","r",stdin); cin >> n; if (isprime(n)){ cout << 1; return 0; } if (n % 2 == 0){ cout << 2; } else { if (isprime(n-2)){ cout << 2; } else cout << 3; } return 0; }
标签：答案 eal namespace opened contains sed input nts 分享
原文地址：http://www.cnblogs.com/ToTOrz/p/7294718.html
