Avito Cool Challenge 2018-B. Farewell Party(思维）

最新推荐文章于 2019-05-25 16:30:16 发布

转载最新推荐文章于 2019-05-25 16:30:16 发布 · 158 阅读

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原文链接：http://www.cnblogs.com/Staceyacm/p/10781914.html

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#数据结构与算法

本文探讨了一道有趣的算法题目，即如何根据派对参与者关于帽子类型的陈述，找出一种可能的帽子分配方案，确保每个人的说法都合理。文章通过示例解释了问题背景，并提供了一段C++代码作为解决方案，展示了如何通过分组和计数来确定每种帽子类型的分配。

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time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Chouti and his classmates are going to the university soon. To say goodbye to each other, the class has planned a big farewell party in which classmates, teachers and parents sang and danced.

Chouti remembered that nn persons took part in that party. To make the party funnier, each person wore one hat among nn kinds of weird hats numbered 1,2,…n1,2,…n. It is possible that several persons wore hats of the same kind. Some kinds of hats can remain unclaimed by anyone.

After the party, the ii-th person said that there were aiai persons wearing a hat differing from his own.

It has been some days, so Chouti forgot all about others' hats, but he is curious about that. Let bibi be the number of hat type the ii-th person was wearing, Chouti wants you to find any possible b1,b2,…,bnb1,b2,…,bn that doesn't contradict with any person's statement. Because some persons might have a poor memory, there could be no solution at all.

Input

The first line contains a single integer nn (1≤n≤1051≤n≤105), the number of persons in the party.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤n−10≤ai≤n−1), the statements of people.

Output

If there is no solution, print a single line "Impossible".

Otherwise, print "Possible" and then nn integers b1,b2,…,bnb1,b2,…,bn (1≤bi≤n1≤bi≤n).

If there are multiple answers, print any of them.

Examples

input

Copy

3
0 0 0

output

Copy

Possible
1 1 1

input

Copy

5
3 3 2 2 2

output

Copy

Possible
1 1 2 2 2

input

Copy

4
0 1 2 3

output

Copy

Impossible

Note

In the answer to the first example, all hats are the same, so every person will say that there were no persons wearing a hat different from kind 11.

In the answer to the second example, the first and the second person wore the hat with type 11 and all other wore a hat of type 22.

So the first two persons will say there were three persons with hats differing from their own. Similarly, three last persons will say there were two persons wearing a hat different from their own.

In the third example, it can be shown that no solution exists.

In the first and the second example, other possible configurations are possible.

题解:看分了几组，假如说有很多人说了有a个人和自己不同，那么就有n－a个人和自己是一组。

代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring> 

using namespace std;

struct node
{
	int id,val;
}p[100005];

bool cmp(node x,node y)
{
	return x.val<y.val;
} 

int ans[100005];
int main()
{
	int n;
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		p[i].id=i;
		scanf("%d",&p[i].val);
	}
	sort(p+1,p+n+1,cmp);
	//for(int t=1;t<=n;t++)
	//{
		//cout<<p[t].val<<" "<<p[t].id<<endl;
	//}
	int last = 1;
	long long int sum=0;
	int cnt=0;
	while(1)
	{
		
		int t=last+(n-p[last].val)-1;
		if(p[t].val==p[last].val)
		{
			cnt++;
			for(int j=last;j<=t;j++)
			{
				ans[p[j].id]=cnt;
			}
			sum+=(n-p[last].val);
			last=t+1;
			if(t==n)break;
		}
		else{
			printf("Impossible\n");
			return 0;
		} 
	}
	if(sum!=n)
	{
		printf("Impossible\n");
	}
	else
	{
		printf("Possible\n");
		for(int i=1;i<=n;i++)
		{
			printf("%d ",ans[i]);
		}
	}
	
	return 0;
}

转载于:https://www.cnblogs.com/Staceyacm/p/10781914.html