Codeforces Round #277 (Div. 2)E题

The next "Data Structures and Algorithms" lesson will be about Longest Increasing Subsequence (LIS for short) of a sequence. For better understanding, Nam decided to learn it a few days before the lesson.

Nam created a sequence a consisting of n (1 ≤ n ≤ 105) elements a1, a2, ..., an (1 ≤ ai ≤ 105). A subsequence ai1, ai2, ..., aik where 1 ≤ i1 < i2 < ... < ik ≤ n is called increasing if ai1 < ai2 < ai3 < ... < aik. An increasing subsequence is called longest if it has maximum length among all increasing subsequences.

Nam realizes that a sequence may have several longest increasing subsequences. Hence, he divides all indexes i (1 ≤ i ≤ n), into three groups:

1. group of all i such that ai belongs to no longest increasing subsequences.
2. group of all i such that ai belongs to at least one but not every longest increasing subsequence.
3. group of all i such that ai belongs to every longest increasing subsequence.

Since the number of longest increasing subsequences of a may be very large, categorizing process is very difficult. Your task is to help him finish this job.

Input

The first line contains the single integer n (1 ≤ n ≤ 105) denoting the number of elements of sequence a.

The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 105).

Output

Print a string consisting of n characters. i-th character should be '1', '2' or '3' depending on which group among listed above index ibelongs to.

Sample test(s)

input1

1

4

output1

3

input2

4

1 3 2 5

output2

3223

input3

4

1 5 2 3

output3

3133

题目大意：给你一个数字序列，先求最长不小降子序列。然后将序列中的数字分为3类，第一类：ai不属于最长子序列；第二类：ai属于最长子序列但不属于每一个最长子序列；第三类：ai属于每一个最长子序列。

题目分析：先用O(nlogn)求出到每一个数时的最长序列，然后处理分类情况。姿势：逆序扫一遍，记录后面最长子序列为j时的最大的数（放松条件），将属于最长子序列的条件最‘松’化，这样就可以将第一类和二三类分开。然后在扫一遍记录属于二三类的数，将长度唯一（长度为j只有这一个数）的记录下来，这些介绍关键路径了（第三类）。

#include<cstdio>
#include<cstring>
int a[100100],f[100100],e[100100],g[100100],h[100100],ans[100100];
int find_mid(int l,int r, int a)
{
    int mid;
    if (r-l<2) return l;
    mid=(l+r)/2;
    if (h[mid]>a) return find_mid(l,mid,a);
    else return find_mid(mid,r,a);
}
int main()
{
    int i,k,n,mid,maxx;
    scanf("%d", &n);
    for (i=1;i<=n;i++) ans[i]=1;
    memset(g,0,sizeof(g));
    memset(f,0,sizeof(f));
    memset(e,0,sizeof(e));
    for (i=1;i<=n;i++) scanf("%d", &a[i]);
    k=1;f[1]=1;h[1]=a[1];
    for (i=2;i<=n;i++)
    if (a[i]>h[k]) {
        k++;f[i]=k;
        h[k]=a[i];
    } else {
        if (a[i]==h[k]) f[i]=k;
        else {
            mid=find_mid(1,k,a[i]);
            if (a[i]==h[mid]||a[i]<h[mid]) {
                f[i]=mid;h[mid]=a[i];
            } else {
                f[i]=mid+1;h[mid+1]=a[i];
            }
        }
    }
    maxx=0;
    for (i=1;i<=n;i++)
    if (f[i]>maxx) maxx=f[i];
    e[maxx+1]=200000;
    for (i=n;i>0;i--)
    if (a[i]<e[f[i]+1]) {
        ans[i]=2;
        if (a[i]>e[f[i]]) e[f[i]]=a[i];
    }
    for (i=1;i<=n;i++)
    if (ans[i]==2) g[f[i]]++;
    for (i=1;i<=n;i++)
    if (ans[i]==2&&g[f[i]]==1) ans[i]=3;
    for (i=1;i<=n;i++) printf("%d",ans[i]);
    printf("\n");
    return 0;
}