Sherlock and his girlfriend
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 1and 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.
Examples
Input
3
Output
2 1 1 2
Input
4
Output
2 2 1 1 2
Note
In the first input, the colors for first, second and third pieces of jewelry having respective prices 2, 3 and 4 are 1, 1 and 2respectively.
In this case, as 2 is a prime divisor of 4, colors of jewelry having prices 2 and 4 must be distinct.
题意：给定一个整数 n, 代表有 n 个珠宝，第 i 个珠宝的价值为 i + 1 . 然后需要给珠宝上颜色，如果一个珠宝的价格 是 另一个价格的 质因数，则这两个珠宝需要上不同的颜色， 问最后 最少需要多少种颜色 k ，输出 这 n 个珠宝的颜色(1 ~ k) .
思路：先素数筛选打表1 到 100000的质数， 最后再用 素数来筛选 求颜色。
AC代码：
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int prime[maxn];
int vis[maxn];
int main()
{
int n;
while(~scanf("%d",&n)){
for(int i = 2;i <= n + 1;i ++){
if(prime[i]) continue;
for(int j = i + i;j <= n + 1;j += i)
prime[j] = 1;
}
for(int i = 1;i <= n + 1;i ++) vis[i] = 1; ///起始都赋颜色为 1
for(int i = 2;i <= n + 1;i ++){
if(prime[i]) continue;
for(int j = i + i;j <= n + 1;j += i){ ///用素数 来 筛选含有此素数为因数的 数，涂上另一种颜色
vis[j] = vis[i] + 1;
}
}
set<int> s;
for(int i = 2;i <= n + 1;i ++)
s.insert(vis[i]);
printf("%d\n",s.size());
for(int i = 2;i <= n + 1;i ++)
printf("%d ",vis[i]);
printf("\n");
}
return 0;
}