Codeforces Round #382 (Div. 1) B. Taxes

B. Taxes
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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 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.

Examples
input
4
output
2
input
27
output
3

题意：要根据自己的收入交税，税额是收入n除本身外的最大因数，可以将收入n分成k份，每份不能为1，分成k份后取每份的因数加和作为税额。

题解：根据素数的因子是除本身外就是1，因此素数收入的税额是最小的，根据哥德巴赫猜想，任何偶数都可拆分成两个质数，因此偶数收入的税额必定是2，素数税额必定是1，奇数分为3中情况，一种作为素数税额为1，一种作为奇数，但减2后得到一个素数，因此可以看成是两个素数之和，税额也为2。第三种是奇数减2之后仍是一个普通奇数，但再减2之后可以得到一个素数，因此是2+2+素数，三个素数之和税额为3。

#include<stdio.h>///判断素数的方法，就是从2一直到sqrt(n)判断取摸是否有为0#include<string.h>///可以不用素数筛，单个数判断素数#include<math.h>bool judge(long long n)///分3种情况，根据哥德巴赫猜想，一个不为2的偶数必能拆成两个素数之和，所以偶数n纳税都是2{    int x=n,i;    for(i=2; i<=sqrt(n); i++)///奇数分两种情况，一种就是素数，另一种是，奇数-2后是一个素数，此时可分成2+一个素数，纳税为2    {        while(x%i==0)        {            return false;///第三种情况，奇数-2不为素数，此时奇数被分为3个素数相加，纳税为3        }    }    return true;}int main()///题目要纳的税是收入n的最大因子(除去n)，并且可以把n分成不为1的很多份{///求纳税最小值    int i,n;    while(scanf("%d",&n)!=EOF)///也就是当分成很多个素数时，纳税都是1    {        if(judge(n))///若是质数。最大因子为1        {            printf("1\n");            continue;        }        if(n%2==0)///若是偶数，可分解为两个质数之和            printf("2\n");        else///非素数的奇数，有两种情况        {            if(judge(n-2))                printf("2\n");            else                printf("3\n");        }    }}///以下是分解质因数的方法//void zhiyinshu(int n)//{//    int x=n,i;//    printf("%d=",n);//    for(i=2;i<=sqrt(n);i++)//    {//        while(x%i==0)//        {//            if(x==i)//                break;//            x/=i;//            printf("%d*",i);//        }//    }//    printf("%d\n",x);//}