CUC-SUMMER-CONTEST-2-I

I - Inversion

HDU-6098

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

Limits
T≤20T≤20
2≤n≤1000002≤n≤100000
1≤Ai≤10000000001≤Ai≤1000000000
∑n≤700000∑n≤700000
Output
For each test case output one line contains n-1 integers, separated by space, ith number is Bi+1Bi+1.
Sample Input
2
4
1 2 3 4
4
1 4 2 3
Sample Output
3 4 3
2 4 4

题意：给你一个序列A，求序列B，B[i]是A[j]的最大值，其中i是j的因子

解法：把A按从大到小排序，然后从前往后取A的值，如果之前的索引是B索引的倍数，则break掉

代码：

#include<iostream>
#include<algorithm>
using namespace std;
int b[100010];
struct node{
    int v,p;
}a[100010];
bool cmp(node x,node y){
    return x.v>y.v;
}
int main()
{
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        for(int i=1;i<=n;i++){
            cin>>a[i].v;
            a[i].p=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].p%i!=0){
                    b[i]=a[j].v;
                    break;
                }
            }
        for(int i=2;i<=n;i++){
            if(i!=2)
                cout<<" ";
            cout<<b[i];
        }
        cout<<endl;
    }
}