題目連結

C. Playing Piano

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Little Paul wants to learn how to play piano. He already has a melody he wants to start with. For simplicity he represented this melody as a sequence a1,a2,…,ana1,a2,…,an of key numbers: the more a number is, the closer it is to the right end of the piano keyboard.

Paul is very clever and knows that the essential thing is to properly assign fingers to notes he's going to play. If he chooses an inconvenient fingering, he will then waste a lot of time trying to learn how to play the melody by these fingers and he will probably not succeed.

Let's denote the fingers of hand by numbers from 11 to 55. We call a fingering any sequence b1,…,bnb1,…,bn of fingers numbers. A fingering is convenient if for all 1≤i≤n−11≤i≤n−1 the following holds:

• if ai<ai+1ai<ai+1 then bi<bi+1bi<bi+1, because otherwise Paul needs to take his hand off the keyboard to play the (i+1)(i+1)-st note;
• if ai>ai+1ai>ai+1 then bi>bi+1bi>bi+1, because of the same;
• if ai=ai+1ai=ai+1 then bi≠bi+1bi≠bi+1, because using the same finger twice in a row is dumb. Please note that there is ≠≠, not == between bibiand bi+1bi+1.

Please provide any convenient fingering or find out that there is none.

Input

The first line contains a single integer nn (1≤n≤1051≤n≤105) denoting the number of notes.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤2⋅1051≤ai≤2⋅105) denoting the positions of notes on the keyboard.

Output

If there is no convenient fingering, print −1−1. Otherwise, print nn numbers b1,b2,…,bnb1,b2,…,bn, each from 11 to 55, denoting a convenient fingering, separated by spaces.

Examples

input

Copy

5
1 1 4 2 2

output

Copy

1 4 5 4 5 

input

Copy

7
1 5 7 8 10 3 1

output

Copy

1 2 3 4 5 4 3 

input

Copy

19
3 3 7 9 8 8 8 8 7 7 7 7 5 3 3 3 3 8 8

output

Copy

1 3 4 5 4 5 4 5 4 5 4 5 4 3 5 4 3 5 4 

Note

The third sample test is kinda "Non stop" song by Reflex.

【題意】

給出一個數列a[n]，讓構造一個滿足下列條件的數列b[n]：如果a[i]>a[i-1]，那麼b[i]>b[i-1]，如果a[i]<a[i-1]，那麼b[i]<b[i-1]，如果a[i]==a[i-1],那麼b[i]!=b[i-1]。

【解題思路】

dp版本：dp[i][j]表示第i個位置是否可以放第j個數字，如果可以dp[i][j]標記為1，並且用pre陣列記錄之前一個位置的數字，最後逆序記錄陣列b即為所求。

記憶化搜尋：直接搜就可以了，沒什麼特別需要注意的。

【程式碼】

dp版本：

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+5;
int dp[maxn][6],a[maxn],pre[maxn][6],b[maxn];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    for(int i=1;i<=5;i++)
        dp[1][i]=1;
    for(int i=1;i<n;i++)
    {
        for(int j=1;j<=5;j++)
        {
            if(!dp[i][j])continue;
            if(a[i]<a[i+1])
            {
                for(int k=j+1;k<=5;k++)
                {
                    dp[i+1][k]=1;
                    pre[i+1][k]=j;
                }
            }
            else if(a[i]>a[i+1])
            {
                for(int k=j-1;k>=0;k--)
                {
                    dp[i+1][k]=1;
                    pre[i+1][k]=j;
                }
            }
            else
            {
                for(int k=1;k<=5;k++)
                {
                    if(k!=j)
                    {
                        dp[i+1][k]=1;
                        pre[i+1][k]=j;
                    }
                }
            }
        }
    }
    int flag=0;
    for(int i=1;i<=5;i++)
    {
        if(dp[n][i]==1)
        {
            int u=i;
            for(int j=n;j>=1;j--)
            {
                b[j]=u;
                u=pre[j][u];
            }
            flag=1;
        }
    }
    if(!flag)printf("-1\n");
    else
    {
        printf("%d",b[1]);
        for(int i=2;i<=n;i++)printf(" %d",b[i]);
        printf("\n");
    }
}

記憶化搜尋：

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+5;
int a[maxn],b[maxn],vis[maxn][6];
int n;
bool dfs(int pos)
{
    if(pos==n)return true;
    for(int i=1;i<=5;i++)
    {
        if(!vis[pos+1][i])
        {
            if((a[pos]==a[pos+1] && b[pos]!=i) || (a[pos]<a[pos+1] && b[pos]<i)
               || a[pos]>a[pos+1] && b[pos]>i)
            {
                b[pos+1]=i;
                vis[pos+1][i]=1;
                if(dfs(pos+1))return true;
            }
        }
    }
    return false;
}
int main()
{
    int flag=0;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    for(int i=1;i<=5;i++)
    {
        b[1]=i;
        if(dfs(1))
        {
            flag=1;
            break;
        }
    }
    if(flag)
    {
        printf("%d",b[1]);
        for(int i=2;i<=n;i++)
            printf(" %d",b[i]);
        printf("\n");
    }
    else printf("-1\n");
}