B. Little Pony and Harmony Chest

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Princess Twilight went to Celestia and Luna's old castle to research the chest from the Elements of Harmony.

A sequence of positive integers bi is harmony if and only if for every two elements of the sequence their greatest common divisor equals 1. According to an ancient book, the key of the chest is a harmony sequence bi which minimizes the following expression:

You are given sequence ai, help Princess Twilight to find the key.

Input

The first line contains an integer n (1 ≤ n ≤ 100) — the number of elements of the sequences a and b. The next line contains n integersa1, a2, ..., an (1 ≤ ai ≤ 30).

Output

Output the key — sequence bi that minimizes the sum described above. If there are multiple optimal sequences, you can output any of them.

Sample test(s)

input

5
1 1 1 1 1

output

1 1 1 1 1 

input

5
1 6 4 2 8

output

1 5 3 1 8 

假想把数列改成1，1，1... 1 满足题意

因此我们对于数列中任意1个数，只要寻找比改成1更优的策略

由于ai≤30，bi≤30+29=59,因此涉及的质数≤59

故可用记忆化搜索+位运算Dp求解

#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 (100+10)
#define MAXAi (30)
#define MAXBi (59)
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,a[MAXN],prime[16] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53};
int dp[1<<16][MAXN],divv[MAXN];
int dfs(int mask,int siz)
{
	if (siz==n+1) return 0;
	int &ret=dp[mask][siz];
	if (ret!=-1) return ret;ret=INF;
	For(i,2*a[siz]-1)
	{
		if (mask&divv[i]) continue;
		ret=min(ret,dfs(mask|divv[i],siz+1)+abs(i-a[siz]));
	}
	return ret;
}
int ans[MAXN];
void find(int mask,int siz)
{
	if (siz==n+1) return ;
	int ret=dp[mask][siz];
	For(i,2*a[siz]-1)
	{
		if (mask&divv[i]) continue; //每个质数最多出现在1个bi因子中
		if (ret==dfs(mask|divv[i],siz+1)+abs(i-a[siz])) 
		{
			ans[siz]=i;
			find(mask|divv[i],siz+1); 
			return;
		}
	}
}
int main()
{
//	freopen("Harmony Chest.in","r",stdin);
//	freopen(".out","w",stdout);
	scanf("%d",&n);
	For(i,n) scanf("%d",&a[i]);
	dfs(0,1);
	MEM(divv)
	memset(dp,-1,sizeof dp);
	For(i,MAXBi)
		Rep(j,16)
		{
			if (i%prime[j]==0) divv[i]|=1<<j;
		}
	dfs(0,1);
	find(0,1);
	
	For(i,n-1) printf("%d ",ans[i]);printf("%d\n",ans[n]);
	
	return 0;
}