Abelian Period

題目連接：

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

Hint

題意

設(shè)S是一個(gè)數(shù)字串，定義函數(shù)occ(S,x)表示S中數(shù)字x的出現(xiàn)次數(shù)。

例如：S=(1,2,2,1,3),occ(S,1)=2,occ(S,2)=2,occ(S,3)=1。

如果對(duì)于任意的i，都有occ(u,i)=occ(w,i)，那么我們認(rèn)為數(shù)字串u和w匹配。

例如：(1,2,2,1,3)≈(1,3,2,1,2)

對(duì)于一個(gè)數(shù)字串S和一個(gè)正整數(shù)k，如果S可以分成若干個(gè)長(zhǎng)度kk的連續(xù)子串，且這些子串兩兩匹配，那么我們稱(chēng)k是串S的一個(gè)完全阿貝爾周期。

給定一個(gè)數(shù)字串S，請(qǐng)找出它所有的完全阿貝爾周期。

題解：

枚舉k，首先k必須是n的約數(shù)，然后就能算出每個(gè)數(shù)字應(yīng)該出現(xiàn)多少次，O(n)檢驗(yàn)即可。

代碼

#include<bits/stdc++.h> using namespace std; const int maxn = 1e5+7; int ok[maxn],a[maxn],c1[maxn],c2[maxn],n; bool check(int x) {memset(c1,0,sizeof(c1));memset(c2,0,sizeof(c2));for(int i=1;i<=x;i++)c1[a[i]]++;for(int i=2;i<=n/x;i++){int l=(i-1)*(x)+1;int r=i*(x);for(int j=l;j<=r;j++){c2[a[j]]++;if(c2[a[j]]>c1[a[j]])return 0;}for(int j=l;j<=r;j++)c2[a[j]]--;}return 1; } void solve() {scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",&a[i]);memset(ok,0,sizeof(ok));for(int i=1;i<=n;i++){if(n%i==0&&!ok[i]){if(check(i)){for(int j=i;j<=n;j+=i)if(n%j==0)ok[j]=1;}}}int fi=0;for(int i=1;i<=n;i++){if(ok[i]){if(fi==0)printf("%d",i),fi=1;else printf(" %d",i);}}printf("\n"); } int main() {int t;scanf("%d",&t);while(t--)solve();return 0; }

轉(zhuǎn)載于:https://www.cnblogs.com/qscqesze/p/5927590.html

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