HDU 5908 Abelian Period (BestCoder Round #88 模拟+暴力)

题目链接http://acm.hdu.edu.cn/showproblem.php?pid=5908

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

题意：

给你n个数字，首先你可以以没k个为一段，要求n分成的各个段中数字的数量和个数都相同。

题解：

首先枚举每个可以被n整除的段，然后再判断是否可以即可，枚举sqrt(n)或者n/2暴力都可以。

判断的方法:首先我们先统计每个元素的个数，对于k个元素一组，因为是均分的，所以每组出现多少个就确定了。现在我们从开始往后面扫，首先没k个一组，多少组就确定了，那么我们现在看，在每一段该元素都有每个数字的出现上限。如果超过了那么必然是错误的k划分，同时我们记录下到达该数字上限的数的个数，和之前的统计如果不同返回false。最终返回true，这样就求出。

代码：

#include <bits/stdc++.h>

using namespace std;

const int maxn = 100000+100;

int s[maxn+10];

int rec[maxn+10];

bool fuck[maxn+10];

int db[maxn+10];

int ans[maxn+10];

int n;

int tol = 0;

inline bool sol(int k)

{

    int tt = n / k;

    memset(db,0,sizeof db);

    for (int i = 0; i < tt; i++){

        int temp = 0;

        for (int j = 1; j <= k; j++){

            db[s[i*k+j]]++;

            if (db[s[i*k+j]]*tt > rec[s[i*k+j]]*(i+1))

                return false;

            else if (db[s[i*k+j]]*tt == rec[s[i*k+j]]*(i+1))

                temp++;

        }

        if (temp != tol)

            return false;

    }

    return true;

}

int main()

{

    int t;

    scanf("%d",&t);

    while (t--){

        tol = 0;

        int ta = 0;

        memset(rec,0,sizeof rec) ;

        scanf("%d",&n);

        for (int i = 1; i <= n; i++){

            scanf("%d",&s[i]);

            rec[s[i]]++;

        }

        for (int i = 1; i <= maxn; i++)

            if (rec[i] != 0)

               tol++;

        for (int i = 1; i*2 <= n; i++){

            if (n%i == 0){

                if (sol(i))

                    printf("%d ",i);

            }

        }

        printf("%d\n",n);

    }

    return 0;

}