cf449C Jzzhu and Apples

最新推荐文章于 2020-07-22 10:56:01 发布

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原文链接：http://www.cnblogs.com/zhber/p/7283775.html

本文解析了一道关于将编号苹果进行最优分组的问题。通过枚举质数及其倍数来实现苹果的有效配对，确保每组苹果编号的最大公约数大于1。文中提供了解决方案并附带代码实现。

Jzzhu has picked n apples from his big apple tree. All the apples are numbered from 1 to n. Now he wants to sell them to an apple store.

Jzzhu will pack his apples into groups and then sell them. Each group must contain two apples, and the greatest common divisor of numbers of the apples in each group must be greater than 1. Of course, each apple can be part of at most one group.

Jzzhu wonders how to get the maximum possible number of groups. Can you help him?

Input

A single integer n (1 ≤ n ≤ 10⁵), the number of the apples.

Output

The first line must contain a single integer m, representing the maximum number of groups he can get. Each of the next m lines must contain two integers — the numbers of apples in the current group.

If there are several optimal answers you can print any of them.

Examples

Input

Output

2
6 3
2 4

Input

Output

Input

Output

瞎鸡儿构造题是坠猥琐的

枚举 除了2以外 的每个质数p，然后枚举它的倍数，看看其中还没被找过的数有多少个。找完的要用什么东西标记下，防止15=3*5这样的被3，5找两次

如果是偶数个，刚好配好了。如果是奇数个，那么把2p提出来，其他的配对。这样提出来好多个2p[i]，他们两两之间都有gcd=2，都可以任意配对。

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<algorithm>
 6 #include<cmath>
 7 #include<queue>
 8 #include<deque>
 9 #include<set>
10 #include<map>
11 #include<ctime>
12 #define LL long long
13 #define inf 0x7ffffff
14 #define pa pair<int,int>
15 #define mkp(a,b) make_pair(a,b)
16 #define pi 3.1415926535897932384626433832795028841971
17 using namespace std;
18 inline LL read()
19 {
20     LL x=0,f=1;char ch=getchar();
21     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
22     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
23     return x*f;
24 }
25 int n;
26 bool mrk[300010];
27 int res[300010],len;
28 int ans1[300010],ans2[300010],ans;
29 int main()
30 {
31     n=read();
32     for (int i=3;i<=n;i++)
33     {
34         if (!mrk[i]&&(i&1))
35         {
36             int rec=i;
37             for (int j=3*i;j<=n;j+=i)
38             {
39                 if (!mrk[j])
40                 {
41                     if (rec){mrk[rec]=mrk[j]=1;ans1[++ans]=rec;ans2[ans]=j;rec=0;}
42                     else rec=j;
43                 }
44             }
45             if (rec&&!mrk[2*i]&&2*i<=n){mrk[rec]=mrk[2*i]=1;ans1[++ans]=rec;ans2[ans]=2*i;rec=0;}
46             else if (2*i<=n)res[++len]=2*i,mrk[2*i]=1;
47         }
48     }
49     for (int i=2;i<=n;i+=2)
50     {
51         if (!mrk[i])res[++len]=i;
52     }
53     if (len&1)len--;
54     for (int i=1;i<=len;i+=2)ans1[++ans]=res[i],ans2[ans]=res[i+1];
55     printf("%d\n",ans);
56     for (int i=1;i<=ans;i++)printf("%d %d\n",ans1[i],ans2[i]);
57 }