CodeForces

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

这个题确实是想到了哥德巴赫….
然而第一次还是wa看了看测试数据
发现忘记了一种情况…

总结一下一个数用哥德巴赫分解完的各种情况

n<4的时候都只能分解出一个质数就是它本身
n>4
偶数全部可以分解成两个质数相加
奇数分两种
第一种可以分成两个质数相加
这只有一种情况比如该数=某质数+2
第二种就是偶数+1
可以分解成两个质数相加+1

#include<iostream>#include<algorithm>#include<map>#include<cstdio>#include<cmath>using namespace std;int ss(long long q){    if (q == 2||q==3||q==5||q==7)return 1;    if (q <= 10)return 0;    for (int a = 3; a*a <= q; a+=2)    {        if (q%a == 0)return 0;    }    return 1;}int main(){#define int long long     int n, q, w;    cin >> n;    if (n > 3)    {        if (n % 2 == 0)        {            cout << 2;            return 0;        }        if (ss(n))        {            cout << 1 << endl;            return 0;        }        if (ss(n - 2))        {            cout << 2 << endl;            return 0;        }        cout << 3 << endl;        return 0;    }    cout << 1;}