Abelian Period

Problem Description

Let S be a number string, and occ(S,x) means the times that number x occurs in S.

i.e. S=(1,2,2,1,3),occ(S,1)=2,occ(S,2)=2,occ(S,3)=1.

String u,w are matched if for each number i, occ(u,i)=occ(w,i) always holds.

i.e. (1,2,2,1,3)≈(1,3,2,1,2).

Let S be a string. An integer k is a full Abelian period of S if S can be partitioned into several continous substrings of length k, and all of these substrings are matched with each other.

Now given a string S, please find all of the numbers k that k is a full Abelian period of S.

 

Input

The first line of the input contains an integer T(1≤T≤10), denoting the number of test cases.

In each test case, the first line of the input contains an integer n(n≤100000), denoting the length of the string.

The second line of the input contains n integers S1,S2,S3,...,Sn(1≤Si≤n), denoting the elements of the string.

 

Output

For each test case, print a line with several integers, denoting all of the number k. You should print them in increasing order.

 

Sample Input

265 4 4 4 5 486 5 6 5 6 5 5 6

 

Sample Output

3 62 4 8

 

Source

BestCoder Round #88

思路： 易知答案一定为n的因子，又因为n不是特别大，一一判断就行了；

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int a[110000],vis[110000];int temp[110000],temp1[110000];int n;int judge(int k){int cnt=0,i;for(i=0;i<k;i++)temp[cnt++]=a[i];sort(temp,temp+k);cnt=0;for(i=k;i<n;i++){temp1[cnt++]=a[i];if(cnt==k){sort(temp1,temp1+k);for(int j=0;j<k;j++){if(temp[j]==temp1[j])        continue;elsereturn 0;}cnt=0;}}return 1;}int main(){int t,i;scanf("%d",&t);while(t--){scanf("%d",&n);for(i=0;i<n;i++){scanf("%d",&a[i]);//vis[a[i]]++;}int f=0;for(i=n;i>=1;i--){if(n%i==0){if(judge(n/i)==1){ if(f==0) printf("%d",n/i); else printf(" %d",n/i); f++;}}}printf("\n");}return 0;}