Codeforces Round #382 (Div. 2) D. Taxes （哥德巴赫猜想）
題目連結
題面
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.
輸入
The first line of the input contains a single integer n (2 ≤ n ≤ 2·109) — the total year income of mr. Funt.
輸出
Print one integer — minimum possible number of burles that mr. Funt has to pay as a tax.
樣例輸入
4
樣例輸出
2
【題意】：某個人有n元的工資，但是他要交稅，交稅的數目就是n的最大因子。比如說6，他的因子有1,2，3；最大的是3，所以要交3元的稅，但是這個人想偷稅漏稅（作為共產主義接班人的我們不能向他學習）。偷稅漏稅的方法就是他把這n元分成幾部分，比如把6分成 3 和 3，那麼他就只交2元稅就可以了。但是不能這幾部分都不能為1， 否則會被發現的。思想：儘量都拆成素數，這樣每部分只要交1，但是炒成的素數還儘量少。這就是哥德巴赫猜想
如果一個數是偶數（2除外）那麼他能分解為兩個質數的和；
如果一個數是奇數那麼它有三種情況：
（1）本身是質數
（2）這個數減二是質數，那麼他就能分解為兩個質數（很顯然就是n-2 和 2）
（3）可以分解為一個質數和一個偶數，就是三個質數
#include<iostream>
using namespace std;
int main(){
int n;
while(cin >> n){
int f = 0, ff = 0;
for(int i = 2; i * i <= n; i++){
if(n % i == 0) f = 1;
if((n - 2) % i == 0) ff = 1;
}
if(f == 1 && n % 2 == 0) {
cout << 2 << endl;
continue;
}
if(f == 0) cout << 1 << endl;
else if(ff == 0) cout << 2 << endl;
else cout << 3 << endl;
}
return 0;
}