CF 337E(Divisor Tree-枚举树节点的父亲)

E. Divisor Tree

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

A divisor tree is a rooted tree that meets the following conditions:

• Each vertex of the tree contains a positive integer number.
• The numbers written in the leaves of the tree are prime numbers.
• For any inner vertex, the number within it is equal to the product of the numbers written in its children.

Manao has n distinct integers a₁, a₂, ..., a_n. He tries to build a divisor tree which contains each of these numbers. That is, for each a_i, there should be at least one vertex in the tree which contains a_i. Manao loves compact style, but his trees are too large. Help Manao determine the minimum possible number of vertices in the divisor tree sought.

Input

The first line contains a single integer n (1 ≤ n ≤ 8). The second line contains n distinct space-separated integers a_i (2 ≤ a_i ≤ 10¹²).

Output

Print a single integer — the minimum number of vertices in the divisor tree that contains each of the numbers a_i.

Sample test(s)

input

26 10

output

7

input

46 72 8 4

output

12

input

17

output

1

Note

Sample 1. The smallest divisor tree looks this way:

Sample 2. In this case you can build the following divisor tree:

Sample 3. Note that the tree can consist of a single vertex.

NOI就因为这个少拿10分了啊。。。现在又因为这个掉Rating了吖囧。。。。

首先明确一点——O(n!) 在n≤10时可过，O(2^n)在n≤20时可过。。。

然后枚举树（都懂得。。）

#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#include<functional>#include<iostream>#include<cmath>#include<cctype>#include<ctime>using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Lson (x<<1)#define Rson ((x<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define MAXN (8+10)long long mul(long long a,long long b){return (a*b)%F;}long long add(long long a,long long b){return (a+b)%F;}long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}typedef long long ll;int n,fa[MAXN]={0},p[MAXN]={0};ll a[MAXN],a2[MAXN],ans=INF;void dfs(int l){if (l==n+1){//For(i,n) cout<<a2[i]<<' ';cout<<endl;//For(i,n) cout<<fa[i]<<' ';cout<<endl;ll tot=0;For(i,n) if (!((!p[i])&&fa[i])) tot++; bool b=0;For(i,n-1) if (!fa[i]) {b=1;break;}tot+=b;For(i,n) tot+=p[i]*!(bool)fa[i];ans=min(ans,tot);//cout<<tot<<endl;/*if (tot==9) {cout<<'d';}*/return;}fa[l]=0;dfs(l+1);Fork(i,l+1,n) if (a2[i]%a[l]==0) fa[l]=i,a2[i]/=a[l],dfs(l+1),a2[i]*=a[l];}int main(){//freopen("Divisor.in","r",stdin);cin>>n;For(i,n) cin>>a[i];sort(a+1,a+1+n);For(i,n) {ll x=a[i];for(ll j=2;j*j<=x;j++)while (x%j==0) p[i]++,x/=j;if (x>1) p[i]++,x=1;if (p[i]==1) p[i]--;}memcpy(a2,a,sizeof(a));dfs(1);cout<<ans<<endl;return 0;}