Codeforces 1375 D. Replace by MEX

You’re given an array of 𝑛 integers between 0 and 𝑛 inclusive.

In one operation, you can choose any element of the array and replace it by the MEX of the elements of the array (which may change after the operation).

For example, if the current array is [0,2,2,1,4], you can choose the second element and replace it by the MEX of the present elements — 3. Array will become [0,3,2,1,4].

You must make the array non-decreasing, using at most 2𝑛 operations.

It can be proven that it is always possible. Please note that you do not have to minimize the number of operations. If there are many solutions, you can print any of them.

–

An array 𝑏[1…𝑛] is non-decreasing if and only if 𝑏1≤𝑏2≤…≤𝑏𝑛.

The MEX (minimum excluded) of an array is the smallest non-negative integer that does not belong to the array. For instance:

The MEX of [2,2,1] is 0, because 0 does not belong to the array.
The MEX of [3,1,0,1] is 2, because 0 and 1 belong to the array, but 2 does not.
The MEX of [0,3,1,2] is 4 because 0, 1, 2 and 3 belong to the array, but 4 does not.
It’s worth mentioning that the MEX of an array of length 𝑛 is always between 0 and 𝑛 inclusive.

Input
The first line contains a single integer 𝑡 (1≤𝑡≤200) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer 𝑛 (3≤𝑛≤1000) — length of the array.

The second line of each test case contains 𝑛 integers 𝑎1,…,𝑎𝑛 (0≤𝑎𝑖≤𝑛) — elements of the array. Note that they don’t have to be distinct.

It is guaranteed that the sum of 𝑛 over all test cases doesn’t exceed 1000.

Output
For each test case, you must output two lines:

The first line must contain a single integer 𝑘 (0≤𝑘≤2𝑛) — the number of operations you perform.

The second line must contain 𝑘 integers 𝑥1,…,𝑥𝑘 (1≤𝑥𝑖≤𝑛), where 𝑥𝑖 is the index chosen for the 𝑖-th operation.

If there are many solutions, you can find any of them. Please remember that it is not required to minimize 𝑘.

Example
inputCopy
5
3
2 2 3
3
2 1 0
7
0 7 3 1 3 7 7
9
2 0 1 1 2 4 4 2 0
9
8 4 7 6 1 2 3 0 5
outputCopy
0

2
3 1
4
2 5 5 4
11
3 8 9 7 8 5 9 6 4 1 2
10
1 8 1 9 5 2 4 6 3 7
Note
In the first test case, the array is already non-decreasing (2≤2≤3).

Explanation of the second test case (the element modified by each operation is colored in red):

𝑎=[2,1,0] ; the initial MEX is 3.
𝑎=[2,1,3] ; the new MEX is 0.
𝑎=[0,1,3] ; the new MEX is 2.
The final array is non-decreasing: 0≤1≤3.
Explanation of the third test case:

𝑎=[0,7,3,1,3,7,7] ; the initial MEX is 2.
𝑎=[0,2,3,1,3,7,7] ; the new MEX is 4.
𝑎=[0,2,3,1,4,7,7] ; the new MEX is 5.
𝑎=[0,2,3,1,5,7,7] ; the new MEX is 4.
𝑎=[0,2,3,4,5,7,7] ; the new MEX is 1.
The final array is non-decreasing: 0≤2≤3≤4≤5≤7≤7.

题意：
0~n的范围，每次可以取MEX（最小的没出现过的数），替换掉一个数。求不大于2*n的操作次数，使得这个序列非递减。

思路：
直接的策略是顺序填，把0全部填完，然后第一个位置填0，依次类推。但是这样会超过2*n的次数。

$n+1$个数填到$n$个格子里，直接填$a[i]=i$。如果当前MEX为$n$，则令$a[n-1]=n$，然后n=n-1，这样之后只需要考虑前$n-1$个格子。

每个位置要么会填上$i$，要么会填上$i+1$，且每种操作只有一次，所以不会超过2*n次操作。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
#include <cmath>
#include <stack>
#include <vector>
 
using namespace std;
 
typedef long long ll;
 
const int maxn = 1e3 + 7;
 
int vis[maxn],a[maxn];
int n;
 
int get() {
    memset(vis,0,sizeof(vis));
    for(int i = 0;i < n;i++) {
        vis[a[i]]++;
    }
    int mex = n;
    for(int i = 0;i < n;i++) {
        if(!vis[i]) {
            return i;
        }
    }
    return mex;
}
 
int main() {
    int T;scanf("%d",&T);
    while(T--) {
        scanf("%d",&n);
        for(int i = 0;i < n;i++) {
            scanf("%d",&a[i]);
        }
        
        vector<int>ans;
        while(n > 0) {
            int mex = get();
            if(mex == n) {
                a[n - 1] = n;
                ans.push_back(n - 1);
                n--;
            } else {
                a[mex] = mex;
                ans.push_back(mex);
            }
        }
        printf("%d\n",ans.size());
        for(int i = 0;i < ans.size();i++) {
            printf("%d ",ans[i] + 1);
        }
        printf("\n");
    }
    return 0;
}