Consecutive Subsequence （DP+map）

标签：forms   rom   math   [1]   输入   sam   rms   lse   self   

You are given an integer array of length nn.

You have to choose some subsequence of this array of maximum length such that this subsequence forms a increasing sequence of consecutive integers. In other words the required sequence should be equal to [x,x+1,…,x+k?1][x,x+1,…,x+k?1] for some value xx and length kk.

Subsequence of an array can be obtained by erasing some (possibly zero) elements from the array. You can erase any elements, not necessarily going successively. The remaining elements preserve their order. For example, for the array [5,3,1,2,4][5,3,1,2,4] the following arrays are subsequences: [3][3], [5,3,1,2,4][5,3,1,2,4], [5,1,4][5,1,4], but the array [1,3][1,3] is not.

Input

The first line of the input containing integer number nn (1≤n≤2?1051≤n≤2?105) — the length of the array. The second line of the input containing nn integer numbers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109) — the array itself.

Output

On the first line print kk — the maximum length of the subsequence of the given array that forms an increasing sequence of consecutive integers.

On the second line print the sequence of the indices of the any maximum length subsequence of the given array that forms an increasing sequence of consecutive integers.

Examples

7
3 3 4 7 5 6 8

4
2 3 5 6 

6
1 3 5 2 4 6

2
1 4 

4
10 9 8 7

1
1 

9
6 7 8 3 4 5 9 10 11

6
1 2 3 7 8 9 

Note

All valid answers for the first example (as sequences of indices):

• [1,3,5,6][1,3,5,6]
• [2,3,5,6][2,3,5,6]

All valid answers for the second example:

• [1,4][1,4]
• [2,5][2,5]
• [3,6][3,6]

All valid answers for the third example:

• [1][1]
• [2][2]
• [3][3]
• [4][4]

All valid answers for the fourth example:

• [1,2,3,7,8,9]

题目大意：

先输入n，然后输入n个数，然后找最长递增子串，输出各个元素的下标。

#include <bits/stdc++.h>
using namespace std;
int a[200005];
map <int,int> ma;
int main()
{
    int n;
    cin>>n;
    for(int i=1,x;i<=n;i++)
        cin>>x,a[i]=x,ma[x]=ma[x-1]+1;
    int pos,maxx=0;
    for(map <int,int>::iterator it=ma.begin();it!=ma.end();it++)
    {
        if(it->second>maxx)
            maxx=it->second,pos=it->first;
    }
    cout<<maxx<<‘\n‘;
    for(int i=1;i<=n;i++)
        if(a[i]==pos-maxx+1)
            cout<<i<<‘ ‘,pos++;
    cout<<‘\n‘;
    return 0;
}

Consecutive Subsequence （DP+map）

标签：forms   rom   math   [1]   输入   sam   rms   lse   self   

原文地址：https://www.cnblogs.com/zdragon1104/p/9319949.html