C. Friends and Gifts

链接:https://codeforces.com/contest/1283/problem/C

There are nn friends who want to give gifts for the New Year to each other. Each friend should give exactly one gift and receive exactly one gift. The friend cannot give the gift to himself.

For each friend the value fifi is known: it is either fi=0fi=0 if the ii-th friend doesn't know whom he wants to give the gift to or 1≤fi≤n1≤fi≤n if the ii-th friend wants to give the gift to the friend fifi.

You want to fill in the unknown values (fi=0fi=0) in such a way that each friend gives exactly one gift and receives exactly one gift and there is no friend who gives the gift to himself. It is guaranteed that the initial information isn't contradictory.

If there are several answers, you can print any.

Input

The first line of the input contains one integer nn (2≤n≤2⋅1052≤n≤2⋅105) — the number of friends.

The second line of the input contains nn integers f1,f2,…,fnf1,f2,…,fn (0≤fi≤n0≤fi≤n, fi≠ifi≠i, all fi≠0fi≠0 are distinct), where fifi is the either fi=0fi=0 if the ii-th friend doesn't know whom he wants to give the gift to or 1≤fi≤n1≤fi≤n if the ii-th friend wants to give the gift to the friend fifi. It is also guaranteed that there is at least two values fi=0fi=0.

Output

Print nn integers nf1,nf2,…,nfnnf1,nf2,…,nfn, where nfinfi should be equal to fifi if fi≠0fi≠0 or the number of friend whom the ii-th friend wants to give the gift to. All values nfinfi should be distinct, nfinfi cannot be equal to ii. Each friend gives exactly one gift and receives exactly one gift and there is no friend who gives the gift to himself.

If there are several answers, you can print any.

Examples

input

Copy

5
5 0 0 2 4

output

Copy

5 3 1 2 4 

input

Copy

7
7 0 0 1 4 0 6

output

Copy

7 3 2 1 4 5 6 

input

Copy

7
7 4 0 3 0 5 1

output

Copy

7 4 2 3 6 5 1 

input

Copy

5
2 1 0 0 0

output

Copy

2 1 4 5 3 

代码:

#include <bits/stdc++.h>
using namespace std;
long long t,n,a,b,k,s;
long long f[200001];
map<long long ,long long >m;
queue<long long>q,p; 
int main()
{
    cin>>n;
    for(int i=1;i<=n;i++)
    {
    	cin>>f[i];
    	if(f[i]!=0)
    	{
    		m[f[i]]=1;
    	}
    	else
    	p.push(i);
    	
	}
    for(int i=1;i<=n;i++)
    {
    	if(m[i]==0)
    	q.push(i);
    }
    for(int i=1;i<=n;i++)
    {
    	if(f[i]==0)
    	{
    		a=q.front();
    		q.pop();
    		if(a==i)
    		{
    			q.push(a);
    			a=q.front();
    			q.pop();
    			if(a==i)
    			{
    				b=p.front();
    				p.pop();
    				if(b==i)
    				{
    					b=p.front();
    					p.pop();
    				}
    				f[i]=f[b];
    				f[b]=a;
    			}
    			else
    			f[i]=a;
    		}
    		else
    		f[i]=a;
    	}
    }
    for(int i=1;i<=n;i++)
    {
    	cout<<f[i]<<" ";
    }
}

 

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07-09 20:50
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码农索隆:1.教育背景和荣誉证书合二为一。 2.获奖项目理一遍,你做了什么,对你求职的岗位有什么帮助,没有就删掉。 3.技能特长和教育背景交换位置。 4.技能特长写的太差,上网上找简历参考。都不用问你别的,一个redis就能把你问住,写写你具体会redis哪些方面的知识。
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