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
【题目链接】:http://codeforces.com/contest/735/problem/D
【题解】
哥德巴赫猜想:
任何一个大于2的偶数都能分解成两个质数的和;
所以；如果输入的n为2；就输出1；
如果大于n；
则判断是否为偶数；为偶数就输出2；
如果为奇数；
①先判断是不是质数，是质数就输出1
②不是质数的话就找离它最近的质数now(now要小于等于n-2)；然后用这个数n减去now;得到差；因为n为奇数、且质数肯定是奇数，所以n-now必然为偶数；
这个时候如果n-now==2，则总的答案为1+1==2,如果n-now!=2，则n-now是大于2的偶数，则n-now可以分解成两个质数的和；则答案为1+2==3；
【完整代码】
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <stack>
#include <string>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
void rel(LL &r)
{
r = 0;
char t = getchar();
while (!isdigit(t) && t!='-') t = getchar();
LL sign = 1;
if (t == '-')sign = -1;
while (!isdigit(t)) t = getchar();
while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
r = r*sign;
}
void rei(int &r)
{
r = 0;
char t = getchar();
while (!isdigit(t)&&t!='-') t = getchar();
int sign = 1;
if (t == '-')sign = -1;
while (!isdigit(t)) t = getchar();
while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
r = r*sign;
}
//const int MAXN = x;
const int dx[5] = {0,1,-1,0,0};
const int dy[5] = {0,0,0,-1,1};
const double pi = acos(-1.0);
int now;
bool is(int x)
{
int ma = sqrt(x);
rep1(i,2,ma)
if (x%i==0)
return false;
return true;
}
int n;
int main()
{
//freopen("F:\\rush.txt","r",stdin);
rei(n);
if (n==2)
{
puts("1");
}
else
{
if (n%2==0)
{
puts("2");
return 0;
}
int now = n;
while (true)
{
if (((n-now>=2) || (n-now==0))&&is(now))
{
n-=now;
break;
}
now--;
}
if (n==0)
cout <<1<<endl;
else
{
int tt = n;
if (tt!=2)
puts("3");
else
puts("2");
}
}
return 0;
}