【hdoj5908】Abelian Period

素颜马尾好姑娘i 2022-07-16 04:45 86阅读 0赞

# Abelian Period #

**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/131072 K (Java/Others)  
Total Submission(s): 602 Accepted Submission(s): 253**

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][BestCoder Round _88]

Recommend

wange2014 | We have carefully selected several similar problems for you: [5921][] [5920][] [5919][] [5918][] [5917][]

题目大意：给你n个数，输出所有的长度使得每段该长度的数字出现的次数相同。

首先，该数必须能被n整除才能出现每个数据段各个数字相等的情况。然后考虑，先统计各个数字出现的总个数，这样每个数字在每个数据段出现的次数唯一确定，写一个判断函数判断是否每个数据段中的数字都等于该唯一确定的数即可。是输出，否跳过。286ms

#include<cstdio>
    #include<iostream>
    #include<cstring>
    #include<cmath>
    #include<algorithm>
    using namespace std;
    const int N = 500000+20;
    int ant[N],rec[N],a[N];
    int tol,n;
    bool judge(int k) {
    	int cnt=n/k;
    	memset(ant,0,sizeof(ant));
    	for(int i=0; i<cnt; i++) {
    		int flag=0;
    		for(int j=1; j<=k; j++) {
    			ant[a[i*k+j]]++;
    			if(ant[a[i*k+j]]*cnt>rec[a[i*k+j]]*(i+1))
    				return false;
    			else if(ant[a[i*k+j]]*cnt==rec[a[i*k+j]]*(i+1))
    				flag++;
    		}
    		if(flag!=tol)
    			return false;
    	}
    	return true;
    }
    int main() {
    	int T;
    	scanf("%d",&T);
    	while(T--) {
    		scanf("%d",&n);
    		memset(rec,0,sizeof(rec));
    		for(int i=1; i<=n; i++) {
    			scanf("%d",&a[i]);
    			rec[a[i]]++;
    		}
    		tol=0;
    		for(int i=1; i<N; i++) {
    			if(rec[i])
    				tol++;
    		}
    		for(int i=1; i*2<=n; i++) {
    			if(n%i==0) {
    			//	printf("--\n");
    				if(judge(i)) {
    					printf("%d ",i);
    				}
    			}
    		}
    		printf("%d\n",n);
    	}
    	return 0;
    }

[BestCoder Round _88]: http://acm.hdu.edu.cn/search.php?field=problem&key=BestCoder+Round+%2388&source=1&searchmode=source
[5921]: http://acm.hdu.edu.cn/showproblem.php?pid=5921
[5920]: http://acm.hdu.edu.cn/showproblem.php?pid=5920
[5919]: http://acm.hdu.edu.cn/showproblem.php?pid=5919
[5918]: http://acm.hdu.edu.cn/showproblem.php?pid=5918
[5917]: http://acm.hdu.edu.cn/showproblem.php?pid=5917