HDU5908-Abelian Period

Abelian Period

                                                                             Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)
                                                                                                          Total Submission(s): 455    Accepted Submission(s): 198

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

题意：设S是一个数字串，函数occ(S,x)occ(S,x)表示S中数字x的出现次数。如果对于任意的i，都有occ(u,i)=occ(w,i)occ(u,i)=occ(w,i)，那么数字串u和w匹配。对于一个数字串S和一个正整数k，如果S可以分成若干个长度为k的连续子串，且这些子串两两匹配，那么k是串S的一个完全阿贝尔周期。给定一个数字串S，找出它所有的完全阿贝尔周期。

#include <iostream>#include <cstring>#include <cstdio>#include <cmath>#include <algorithm>using namespace std;const int INF=0x3f3f3f3f;int main(){    int t,n;    int a[100090],x[100090],b[100090],c[100090];    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(int i=1; i<=n; i++)            scanf("%d",&a[i]);        int sum=0;        for(int i=1; i<=n/2; i++)        {            if(n%i==0)            {                for(int j=1;j<1+i;j++)                    b[j-1]=a[j];                sort(b,b+i);                int k=0,flag=1;                for(int j=i+1;j<=n;j++)                {                    c[k++]=a[j];                    if(k==i)                    {                        sort(c,c+i);                        for(int p=0;p<i;p++)                        {                            if(c[p]!=b[p]) {flag=0;break;}                        }                        k=0;                    }                    if(!flag) break;                }                if(flag) x[sum++]=i;            }        }        for(int i=0; i<sum; i++)            printf("%d ",x[i]);        printf("%d\n",n);    }    return 0;}