F - Insert / Time Limit: 2 sec / Memory Limit: 10...

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問題

F - Insert /
Time Limit: 2 sec / Memory Limit: 1024 MB

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Score :
500 points

Problem Statement
There is an empty array
A. For
i=1,2,…,N, perform the following operation in order:

Insert the number
i into
A so that it becomes the
P
i

-th element from the beginning.
More precisely, replace
A with the concatenation of the first
P
i

−1 elements of
A, then
i, then the remaining elements of
A starting from the
P
i

-th element, in this order.
Output the final array
A after all operations have been completed.

Constraints
1≤N≤5×10
5

1≤P
i

≤i
All input values are integers.
Input
The input is given from Standard Input in the following format:

N
P
1

P
2


P
N

Output
Let the final array be
A=(A
1

,A
2

,…,A
N

). Print
A
1

,A
2

,…,A
N

in this order, separated by spaces.

Sample Input 1
Copy
4
1 1 2 1
Sample Output 1
Copy
4 2 3 1
The operations are performed as follows:

Insert the number
1 so that it becomes the 1st element of
A. Now,
A=(1).
Insert the number
2 so that it becomes the 1st element of
A. Now,
A=(2,1).
Insert the number
3 so that it becomes the 2nd element of
A. Now,
A=(2,3,1).
Insert the number
4 so that it becomes the 1st element of
A. Now,
A=(4,2,3,1).
Sample Input 2
Copy
5
1 2 3 4 5
Sample Output 2
Copy
1 2 3 4 5
Language
Source Code

  • at most 512 KiB
  • Your source code will be saved as Main.extension.

Please give correct c++ code all tests pass please and also give FULL all test pass code please and look constraints and using constratints give optimzed code but all tests pass code please take care of this...!!
write simple code which pass all tests please within the range

思考

Considering efficient insertion

The problem requires constructing an array A by inserting elements at specific positions defined by array P. Since an approach using vector insertion in O(n) would lead to 250 billion operations, I need to find a more efficient solution. One option is using a balanced tree data structure, like a balanced binary search tree (BST), which can support insertions in O(log N) time and handle the problem constraints, especially with N up to 500,000.

Simulating insertion backward

I'm thinking about the best way to insert elements in array A efficiently. A standard approach like using std::list with an iterator might take O(n) for each operation, which isn't ideal. Then there's the idea of using Fenwick tree or segment tree, but what if we flip the process and work backward from the end of the array? By iterating from N down to 1, I can insert each element at position P_i-1 in a vector, simulating the original insertions more efficiently. Let's test this through the example.

Restoring permutation from positions

The goal is to restore the final array from the given positions. For each element, I must place it at the P_i-th available position, starting from the end. This process resembles the "restore permutation" problem, which can be efficiently done with a Fenwick tree, segment tree, or other similar data structures. By simulating the process backward, we can correctly reconstruct the array in O(N log N) time. This approach works well when treating available positions like an ordered list.

Reasoned for 15 seconds

回答

We can solve this problem by "reversing" the insertion process. Instead of simulating the insertions from 1 to NN (which would be too slow for NN up to 5×1055\times10^5), we can compute the final array by “placing” the numbers into the available positions.

Explanation

Imagine you have NN empty slots (positions 11 to NN). Initially, all positions are available. For each ii (from 11 to NN) the operation “insert ii so that it becomes the PiP_i‑th element” means that when ii is inserted, it is placed into the PiP_i-th available slot (if you think about the positions that are still free).

It turns out that if we process the numbers in reverse order (from NN down to 11), we can decide which available slot each number must go to. In other words, for i=N,N1,,1i = N, N-1, \dots, 1:

  1. Choose the PiP_i-th available position.
  2. Place the number ii in that position.
  3. Mark that position as no longer available.

This is a standard “inverse permutation from an inversion sequence” problem.

To quickly get the kk‑th available position we use a Fenwick Tree (Binary Indexed Tree). We initialize the tree with 11 for every available position. Then, finding the kk-th available position can be done in O(logN)O(\log N) time using a binary search over the prefix sums of the tree.

The Code

Below is the complete C++ solution that implements the above idea:


cpp
#include <iostream> #include <vector> using namespace std; // Binary Indexed Tree (Fenwick Tree) for sum queries and point updates. struct BIT { int n; vector<int> tree; BIT(int n) : n(n), tree(n + 1, 0) { } // Increase value at index i by delta. void update(int i, int delta) { for(; i <= n; i += i & -i) tree[i] += delta; } // Compute prefix sum for index i. int sum(int i) { int s = 0; for(; i > 0; i -= i & -i) s += tree[i]; return s; } // Find smallest index such that prefix sum >= value. int lower_bound(int value) { int idx = 0; // For n up to 5e5, 2^19 = 524288 is enough. for (int bit = 1 << 19; bit; bit >>= 1) { int next = idx + bit; if(next <= n && tree[next] < value) { value -= tree[next]; idx = next; } } return idx + 1; } }; int main(){ ios::sync_with_stdio(false); cin.tie(nullptr); int N; cin >> N; // Read P[1] to P[N] (1-indexed) vector<int> P(N + 1); for (int i = 1; i <= N; i++){ cin >> P[i]; } // Initialize BIT: mark all positions as available (value 1). BIT bit(N); for (int i = 1; i <= N; i++){ bit.update(i, 1); } // ans[i] will hold the final number at position i. vector<int> ans(N + 1, 0); // Process numbers from N down to 1. for (int i = N; i >= 1; i--){ // Find the P[i]-th available position. int pos = bit.lower_bound(P[i]); ans[pos] = i; // Place number i in that position. bit.update(pos, -1); // Mark the position as occupied. } // Output the final array. for (int i = 1; i <= N; i++){ cout << ans[i] << (i == N ? "\n" : " "); } return 0; }

How the Code Works

  1. Input Reading:
    We read the integer NN and the PiP_i values. Note that the PP array is stored using 1-indexing for easier matching with our BIT implementation.

  2. BIT Initialization:
    We create a BIT of size NN and initialize all positions with value 11 (indicating that all positions are initially available).

  3. Reverse Processing:
    We iterate from i=Ni = N down to 11. For each ii:

    • We find the PiP_i-th available position using the lower_bound function.
    • Place ii into that position.
    • Update the BIT to mark that position as no longer available.
  4. Output:
    Finally, we output the constructed array.

This solution runs in O(NlogN)O(N \log N) time and uses O(N)O(N) memory, which is efficient given the constraints.

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