CodeForces 569B Inventory 货物编号

原题： http://codeforces.com/contest/569/problem/B

题目：

Inventory

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

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.

Sample test(s)

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.

思路：

对于给定一系列有编号的物品。他们的编号可能反复，也有能够超过了物品总个数。如今要求给全部物品又一次编号，让全部物品编号都在个数范围内。而且不反复，改动次数尽可能小，输出改动后的编号。

代码：

#include <iostream>
#include"stdio.h"
#include"stdlib.h"
#include"string.h"
using namespace std;
const int N = 100005;
int a[N];
bool b[N];  //当前值是否存在
bool c[N];  //当前点的值是否反复
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        memset(a,0,sizeof(a));
        memset(b,false,sizeof(b));
        memset(c,false,sizeof(c));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            if(b[a[i]]==false&&a[i]<=n)
            //对于给定范围内的第一次出现的数我们不改动
                b[a[i]]=true;
            //否则要改动
            else c[i]=true;
        }
        int tt=1;
        for(int i=1;i<=n;i++)
        {
            //对于每一个要改动的点
            if(c[i]==true)
            {
                //找到没有出现的元素
                while(b[tt])    tt++;
                //去替换a[i]的值
                a[i]=tt;
                //并将该元素标记为出现过
                b[tt]=true;
            }
        }
        for(int i=1;i<n;i++)
            printf("%d ",a[i]);
        printf("%d\n",a[n]);
    }

    return 0;
}