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

241 2 3 441 4 2 3

 

Sample Output

3 4 32 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;}}