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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
#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;
}
HDU 5908 Abelian Period(暴力+想法题）
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原文地址：http://www.cnblogs.com/zzy19961112/p/5935723.html
