hdu6098思维+队列

Inversion

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

Problem Description

Give an array A, the index starts from 1.
Now we want to know Bi=maxi∤jAj , i≥2.

 

Input

The first line of the input gives the number of test cases T; T test cases follow.
Each case begins with one line with one integer n : the size of array A.
Next one line contains n integers, separated by space, ith number is Ai.

Limits
T≤20
2≤n≤100000
1≤Ai≤1000000000
∑n≤700000

 

Output

For each test case output one line contains n-1 integers, separated by space, ith number isBi+1.

 

Sample Input

2
4
1 2 3 4
4
1 4 2 3

 

Sample Output

3 4 3
2 4 4

 思路：

将输入数据排序，然后将2到n放入队列，然后不整除的赋值为最大，会被整除的放入另一个队列，搜索完不整除的队列后，把能整除的放入不整除的队列中。

#include<cstdio>
#include<iostream>
#include<queue> 
#include<string.h>
#include<algorithm>
using namespace std;
struct node{
	int id;
	int num;
}a[100010];
bool cmp(node x,node y)
{
	return x.num>y.num;
}
int b[100010];
int main()
{
	queue<int> q1,q2;
	int n;
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		for(int i=0;i<n;i++)
		{
			a[i].id=i+1;
			scanf("%d",&a[i].num);
		}	
	//	for(int i=0;i<n;i++)
	//	cout<<a[i].num<<" ";
	//	cout<<endl;
		sort(a,a+n,cmp);
		for(int i=1;i<=n;i++)
		{
			q1.push(i);
		}
		int k=1,j=0;
		while(j<n&&!q1.empty())
		{
			while(!q1.empty())
			{
				int id=q1.front();
				if(a[j].id%id==0)
				{
					q2.push(id);
				}	
				else
				{
					b[id]=a[j].num;
				}
				q1.pop();
			}
			while(!q2.empty())
			{
					q1.push(q2.front());
					q2.pop();
			}
			j++;
		}
		
		for(int i=2;i<=n;i++)
		{
			printf("%d",b[i]);
			if(i!=n)
			printf(" ");
		}	
		cout<<endl;
	}
}