bestcoder——数组划分

Abelian Period

 
 Accepts: 400
 
 Submissions: 961

 Time Limit: 2000/1000 MS (Java/Others)
 
 Memory Limit: 262144/131072 K (Java/Others)

Problem Description

Let SS be
a number string, and occ(S,x)occ(S,x) means
the times that number xx occurs
in SS.
i.e. S=(1,2,2,1,3),occ(S,1)=2,occ(S,2)=2,occ(S,3)=1S=(1,2,2,1,3),occ(S,1)=2,occ(S,2)=2,occ(S,3)=1.
String u,wu,w are
matched if for each number ii, occ(u,i)=occ(w,i)occ(u,i)=occ(w,i) always
holds.
i.e. (1,2,2,1,3)\approx(1,3,2,1,2)(1,2,2,1,3)≈(1,3,2,1,2).
Let SS be
a string. An integer kk is
a full Abelian period of SS if SS can
be partitioned into several continous substrings of length kk,
and all of these substrings are matched with each other.
Now given a string SS,
please find all of the numbers kk that kk is
a full Abelian period of SS.

Input

The first line of the input contains an integer T(1\leq
T\leq10)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\leq
100000)n(n≤100000),
denoting the length of the string.
The second line of the input contains nn integers S_1,S_2,S_3,...,S_n(1\leq
S_i\leq n)S​1​​,S​2​​,S​3​​,...,S​n​​(1≤S​i​​≤n),
denoting the elements of the string.

Output

For each test case, print a line with several integers, denoting all of the number kk.
You should print them in increasing order.

Sample Input

Copy

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

Sample Output

3 6
2 4 8

Statistic | Submit | Clarifications | Back

先找其所有约数，然后对每一个约数进行逐一判断

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 2009
#define INF 9999999999999
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=5e2+10;
using namespace std;
typedef long long ll;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
ll my_min(ll x,ll y){ return x>y?y:x;};

int main()
{
int t;
int n;
int a[100005];
int b[100005];
while(scanf("%d",&t)!=EOF)
{
for(int k=1;k<=t;k++)
{
int len = 0;
memset(b,0,sizeof(b));
memset(a,0,sizeof(a));
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);

for(int i=1;i<=n-1;i++)//找到约数
{
if(n%i==0)
b[len++] = i;
}
//for(int i=1;i<len;i++)
// printf("%d ",b[i]);
//printf("\n");

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

int t1[100005];
int t2[100005];
int len1 = n/b[i];

int flag ;
flag = 1;
for(int j=1;j<len1;j++)//判断
{

for(int i1 = (j-1)*b[i]+1;i1<=j*b[i];i1++)
{
t1[i1-(j-1)*b[i]] = a[i1];
}
for(int i1 = j*b[i]+1;i1 <= (j+1)*b[i];i1++)
{
t2[i1 - j*b[i]] = a[i1];
}
sort(t1+1,t1+1+b[i]);
sort(t2+1,t2+1+b[i]);
int i1;
for(i1=1;i1<=b[i];i1++)
{
if(t1[i1]!=t2[i1])
{
break;
}
}

if(i1!=b[i]+1)
{
flag = 0;
}
}
if(flag)
printf("%d ",b[i]);
}

printf("%d\n",n);
}
}
return 0;
}