题干：
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
Input
7
3 3 4 7 5 6 8
Output
4
2 3 5 6
Input
6
1 3 5 2 4 6
Output
2
1 4
Input
4
10 9 8 7
Output
1
1
Input
9
6 7 8 3 4 5 9 10 11
Output
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：
找一个最长的子序列，使得这个序列的数字是连续的。要求你输出最长子序列长度，和分别位于原序列中的位置下标，spj。
解题报告：
直接map暴力一下就好。假的dp。
AC代码：
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
int n,ans,mx,a[MAX];
map<int,int> mp;
int main()
{
cin>>n;
for(int x,i = 1; i<=n; i++) {
scanf("%d",&x);a[i]=x;
mp[x] = mp[x-1]+1;
if(mp[x] > ans) {
ans = mp[x];
mx=x;
}
}
cout << ans << endl;
int cur = mx-ans+1;
for(int i = 1; i<=n; i++) {
if(a[i] == cur) printf("%d ",i),cur++;
}
return 0 ;
}
