HDU 5908 Abelian Period(暴力+想法题）

传送门

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

2 
6 
5 4 4 4 5 4 
8 
6 5 6 5 6 5 5 6

Sample Output

3 6 
2 4 8

思路

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

 

+ View Code?

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38

#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; }