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 allof these substrings are matched with each other.
Now given a string S, pleasefind all of the numbers k that k is a full Abelian period of S.

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.

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

题意：将一个数字串分成长度为k的几个连续区间，如果每个区间内各个元素出现的次数相同，则称k为一个阿贝尔周期，从小到大打印所有阿贝尔周期

可以知道，k必然是n的约数，因此统计每个数出现的次数，求出这些次数的最大公约数，枚举这个最大公约数的约数x，即为可分为的m段，k值即为N/x

#include<bits/stdc++.h>using namespace std;const int maxn = 100005;int gcd(int a,int b){    return b?gcd(b,a%b):a;}int main(){    int T;    scanf("%d",&T);    while (T--)    {        int N,a[maxn],cnt[maxn];        memset(cnt,0,sizeof(cnt));        scanf("%d",&N);        for (int i = 0;i < N;i++)        {            scanf("%d",&a[i]);            cnt[a[i]]++;        }        int tmp = cnt[1];        for (int i = 2;i < maxn;i++)    tmp = gcd(tmp,cnt[i]);        bool first = true;        for (int i = tmp;i >= 1;i--)        {            if (tmp % i == 0)            {                first?printf("%d",N/i):printf(" %d",N/i);                first = false;            }        }        printf("\n");    }    return 0;}