codeforce 849D. Make a Permutation!

Ivan has an array consisting of n elements. Each of the elements is an integer from 1 to n .

Recently Ivan learned about permutations and their lexicographical order. Now he wants to change (replace) minimum number of elements in his array in such a way that his array becomes a permutation (i.e. each of the integers from 1 to n was encountered in his array exactly once). If there are multiple ways to do it he wants to find the lexicographically minimal permutation among them.

Thus minimizing the number of changes has the first priority, lexicographical minimizing has the second priority.

In order to determine which of the two permutations is lexicographically smaller, we compare their first elements. If they are equal — compare the second, and so on. If we have two permutations x and y , then x is lexicographically smaller if x i < y i , where i is the first index in which the permutations x and y differ.

Determine the array Ivan will obtain after performing all the changes.

The first line contains an single integer n (2 ≤ n ≤ 200 000) — the number of elements in Ivan's array.

The second line contains a sequence of integers a 1, a 2, ..., a n (1 ≤ a i ≤ n ) — the description of Ivan's array.

In the first line print q — the minimum number of elements that need to be changed in Ivan's array in order to make his array a permutation. In the second line, print the lexicographically minimal permutation which can be obtained from array with q changes.

In the first example Ivan needs to replace number three in position 1 with number one, and number two in position 3 with number four. Then he will get a permutation [1, 2, 4, 3] with only two changed numbers — this permutation is lexicographically minimal among all suitable.

In the second example Ivan does not need to change anything because his array already is a permutation.

题意：给以一个序列，问最少改变多次次可以得到一个字典序最小的全排列。

题解：看到时间和这个数量级就应该想到贪心，核心就是需要改变的数字一定是重复的数字。我们把所有重复数字标记一下，然后把需要加进去的数字尽可能的放在靠前的位置（如果要加入的数字比重复数字小的话），重复的数字也要留一个位置，这里特殊处理一下，如果要加进去的数字比重复数字小，那么这个位置的重复数字就不要变动（说的有点绕，看下代码就好了）

ac代码：

#include <cstdio>

#include <cstring>

#include <iostream>

#include <queue>

#include <algorithm>

#include <cmath>

#include <map>

using namespace std;

int vis[];

int add[];

//int zzvis[200010];

int a[];

int mark[];

struct node

{

    int key;

    int pos;

}ff[];

int main()

{

    int n;

    cin>>n;

    memset(vis,,sizeof(vis));

    memset(mark,,sizeof(mark));

   // memset(zzvis,0,sizeof(zzvis));

    int fcnt=;

    for(int i=;i<n;i++)

    {

        scanf("%d",&a[i]);

        vis[a[i]]++;

    }

    int cnt=;// q

    for(int i=;i<=n;i++)

    {

        if(vis[i]==) add[cnt++]=i;

        if(vis[a[i-]]>=)

        {

            ff[fcnt].key=a[i-];

            ff[fcnt++].pos=i-;

        }

    }

    int now=;

    for(int i=;i<cnt;i++)

    {

       // if(i==cnt-1) cout<<now<<endl;

        for(int j=now;j<fcnt;j++)

        {

            node temp=ff[j];

            if(vis[temp.key]==) // 当这个数字只有一个的时候

            {

                if(mark[temp.key]==) // 如果数字已经留了一次，就直接替换

                {

                   a[temp.pos]=add[i];

                   now=j+;

                   break;

                }

                else continue; // 不然这个位置就不要变动了

            }

            if(temp.key < add[i] && mark[temp.key]==) // 如果要加入的数字比当前遍历到的重复数字小 且这个重复数字还没定位置的时候 定下位置

            {

                vis[temp.key]--;

                mark[temp.key]=;

               // cout<<"123"<<endl;

                continue;

            }

            else // 直接替换

            {

                a[temp.pos]=add[i];

                vis[temp.key]--;

                now=j+;

                break;

            }

        }

    }

    //

    cout<<cnt<<endl;

    cout<<a[];

    for(int i=;i<n;i++)

    {

        cout<<' '<<a[i];

    }

    cout<<endl;

    return ;

}