hdu 6098

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 is Bi+1.

Sample Input
2
4
1 2 3 4
4
1 4 2 3

Sample Output
3 4 3
2 4 4

题解：

多校的题有点难理解啊。
i∤j’表示j%i!=0
排序优化一下，以数据大小排序，同时记录索引值，从2开始枚举，计算b数组的值，O(n)扫排序后的a数组，当遇到第一个j%i!=0，记录即可。

代码：

#include <bits/stdc++.h>using namespace std;typedef long long LL;const int maxn = 100000+10;struct node{    int id;    LL data;};bool cmp(node q,node p){    return q.data>p.data;}int main(){    int T;    cin>>T;    LL n;    node a[maxn];    vector<LL> b;    while(T--)    {        cin>>n;        LL tmp;        for(int i=1;i<=n;i++)        {            scanf("%lld",&tmp);            a[i].data=tmp;            a[i].id=i;        }        sort(a+1,a+n+1,cmp);        for(int i=2;i<=n;i++)        {           for(int j=1;j<=n;j++)           {               if(a[j].id%i!=0)               {                   b.push_back(a[j].data);                   break;               }           }        }        int siz = b.size();        for(int i=0;i<siz-1;i++)        {           printf("%lld ",b[i]);        }        printf("%lld",b[siz-1]);        cout<<endl;        b.clear();    }    return 0;}