HDU5908-Abelian Period

最新推荐文章于 2019-05-22 19:45:24 发布

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最新推荐文章于 2019-05-22 19:45:24 发布

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本文介绍了一种算法，用于寻找给定数字串的所有完全阿贝尔周期。通过将字符串分割成等长子串并比较各子串中数字出现的频率来判断是否构成阿贝尔周期。

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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

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

Sample Output

  
   3 6
2 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;
}