题干：
Companies always have a lot of equipment, furniture and other things. All of them should be tracked. To do this, there is an inventory number assigned with each item. It is much easier to create a database by using those numbers and keep the track of everything.
During an audit, you were surprised to find out that the items are not numbered sequentially, and some items even share the same inventory number! There is an urgent need to fix it. You have chosen to make the numbers of the items sequential, starting with 1. Changing a number is quite a time-consuming process, and you would like to make maximum use of the current numbering.
You have been given information on current inventory numbers for n items in the company. Renumber items so that their inventory numbers form a permutation of numbers from 1 to n by changing the number of as few items as possible. Let us remind you that a set of n numbers forms a permutation if all the numbers are in the range from 1 to n, and no two numbers are equal.
Input
The first line contains a single integer n — the number of items (1 ≤ n ≤ 105).
The second line contains n numbers a1, a2, ..., an (1 ≤ ai ≤ 105) — the initial inventory numbers of the items.
Output
Print n numbers — the final inventory numbers of the items in the order they occur in the input. If there are multiple possible answers, you may print any of them.
Examples
Input
3 1 3 2
Output
1 3 2
Input
4 2 2 3 3
Output
2 1 3 4
Input
1 2
Output
1
Note
In the first test the numeration is already a permutation, so there is no need to change anything.
In the second test there are two pairs of equal numbers, in each pair you need to replace one number.
In the third test you need to replace 2 by 1, as the numbering should start from one.
题目大意：
题意1：有n个物品需要标记，每个物品标号都不能相同，要从1开始，且保证充分利用之前的标记，输出最后每个物品的标记。
题意2：给出一个数字n，接下来是n个数，把其中的重复的数或者大于n的数进行替换，使得整个数列是由1~n来组成的，可能会有多种答案，输出其中任意一种。详情直接看样例。
解题报告：
水题，打个标记处理一下就好了。
AC代码：
#include<bits/stdc++.h>
using namespace std;
const int MAX = 1e5 + 5;
int a[MAX];
bool vis[MAX];
int ans[MAX];
int main()
{
int n;
cin>>n;
memset(ans,-1,sizeof(ans));
memset(vis,0,sizeof(vis));
for(int i = 1 ; i <= n ; i++){
cin >> a[i];
if(!vis[a[i]] && a[i] <= n){
ans[i] = a[i];
vis[a[i]]=1;
}
}
int cur = 1;
for(int i = 1 ; i <= n ; i ++){
if(ans[i] == -1){
for(int j = cur; j<=n ; j++){
if(!vis[j]){
cur = j;
ans[i] =cur;
vis[cur]=1;
cur++;
break;
}
}
}
}
for(int i = 1; i<=n ; i++) {
printf("%d%c",ans[i],i == n ? '\n' : ' ');
}
return 0;
}