Codeforce 338 C. Divisor Tree(爆搜，3级)

C. 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.

思路：直接爆搜吧。

#include<cstring>#include<iostream>#include<cstdio>#include<algorithm>#include<map>#define FOR(i,a,b) for(int i=a;i<=b;++i)#define clr(f,z) memset(f,z,sizeof(f))#define LL long long#define ll(x) (1<<x)using namespace std;LL f[12];int b[12],fa[12],id[12];int n,ans;int get(LL x){ if(x==1)return 0;  int num=0;  for(LL i=2;i*i<=x;++i)  {    while(x%i==0)    {      ++num;x/=i;    }  }  if(x>1)++num;  return num;}void DP(int x)///算非叶子节点数{  if(x==0)  {    int num=0,ddd=0;    FOR(i,1,n)    if(fa[i]==n+1)    {      ++num;ddd+=b[i];    }    if(num!=1)ddd++;///多一个根节点    FOR(i,1,n)if(id[i]==0&&b[i]==1)///叶子    --ddd;    ans=min(ans,ddd+n);    return ;  }  FOR(i,x+1,n+1)///0 除任何数都是0  if(f[i]%f[x]==0)  {     id[i]++;///add a edge    f[i]/=f[x];    fa[x]=i;    DP(x-1);    f[i]*=f[x];    id[i]--;  }}int main(){  while(~scanf("%d",&n))  {    FOR(i,1,n)    {scanf("%lld",&f[i]);    }    sort(f+1,f+n+1);    clr(b,0);clr(fa,-1);    FOR(i,1,n)    {b[i]=get(f[i]);    // cout<<b[i]<<endl;    }    ans=9999999;clr(id,0);    DP(n);    printf("%d\n",ans);  }}

、、、、DP

#include<stdio.h>#include<string.h>#include<algorithm>#include<cstring>#include<iostream>using namespace std;int dp[1100],dp1[1100];__int64 a[11];int arr[11];int main(void){    int n,i,j,k;    cin>>n;    for(int i=0;i<n;i++)    cin>>a[i];    sort(a,a+n);    dp[1]=1;    for(i=1;i<n;i++)    {    memset(dp1,0,sizeof(dp1));    for(j=0;j<(1<<i);j++)if(dp[j])        for(k=j;;k=((k-1)&j))        {        __int64 num=a[i];        int is=0;        for(int t=0;t<i;t++)if(k&(1<<t))        {            if(num%a[t])            {            is=1;            break;            }            num/=a[t];        }        if(!is)            dp1[j-k+(1<<i)]=1;        if(k==0) break;        }    memcpy(dp,dp1,sizeof(dp));    }    __int64 ans=0;    for(i=0;i<n;i++)    {   __int64 num=a[i];   j=0;   for(__int64 t=2;t*t<=num;t++)if(num%t==0)       while(num%t==0)       {               j++;               num/=t;           }   if(num>1) j++;   if(j==1) arr[i]=1,ans++;   else arr[i]=j,ans+=j+1;    }    int mm=2100000000;    for(i=0;i<(1<<n);i++)if(dp[i])    {   k=ans;   int cc=0;   for(j=0;j<n;j++)            if(!(i&(1<<j)))           k-=arr[j];        else           cc++;   if(cc!=1)       k++;   mm=min(mm,k);    }    ans=mm;    cout<<ans<<endl;    return 0;}