CF776B Sherlock and his girlfriend

Description

Sherlock has a new girlfriend (so unlike him!). Valentine's day is coming and he wants to gift her some jewelry.

He bought n pieces of jewelry. The i-th piece has price equal to i + 1, that is, the prices of the jewelry are 2, 3, 4, ... n + 1.

Watson gave Sherlock a challenge to color these jewelry pieces such that two pieces don't have the same color if the price of one piece is a prime divisor of the price of the other piece. Also, Watson asked him to minimize the number of different colors used.

Help Sherlock complete this trivial task.

Input

The only line contains single integer n (1 ≤ n ≤ 100000) — the number of jewelry pieces.

Output

The first line of output should contain a single integer k, the minimum number of colors that can be used to color the pieces of jewelry with the given constraints.

The next line should consist of n space-separated integers (between 1 and k) that specify the color of each piece in the order of increasing price.

If there are multiple ways to color the pieces using k colors, you can output any of them.

 

题意：N个点，标号2....N-1，给这些点染色，要求如果a是b的质因子，则a,b不能同色，求一种颜色数最少的方案。

题解：先预处理出区间内的质数，和合数，很简单如果只有质数，那么谁也不可能是谁的质因子，所以就一种方案，如果是合数那只能与它的“质因子”不同颜色，

所以颜色数至多是2，至少是1，最后输出方案时，把所有质数染成1，把所有合数染成2即可，这是一个二分图。

 

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>

using namespace std;

int is_prime[100005],prime[100005],cnt,n;
bool flag;

inline void ou_shai(int x)
{
    for(int i=2;i<=x;i++)
    {
        if(is_prime[i]==0) is_prime[i]=i,prime[++cnt]=i;
        for(int j=1;j<=cnt;j++)
        {
            if(prime[j]*i>x||prime[j]>is_prime[i]) break;
            flag=true;
            is_prime[prime[j]*i]=prime[j];
        }
    }
}

int main()
{
    scanf("%d",&n);
    ou_shai(n+1);
    if(flag) printf("2\n");
    else printf("1\n");
    for(int i=2;i<=n+1;i++)
       if(is_prime[i]==i) printf("%d ",1);
       else printf("%d ",2);
    return 0;
}