Abelian Period

Accepts: 288

Submissions: 984

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 262144/131072 K (Java/Others)

问题描述

设

输入描述

输入的第一行包含一个正整数 $第二行包含nnn个正整数S1,S2,S3,...,Sn(1≤Si≤n)S_1,S_2,S_3,...,S_n(1leq S_ileq n)S​1​​,S​2​​,S​3​​,...,S​n​​(1≤S​i​​≤n)，表示这个数字串。$

输出描述

对于每组数据，输出一行若干个整数，从小到大输出所有合法的

输入样例

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

输出样例

3 6
2 4 8

思路：枚举n的因子，然后我们去检验这个是否符合就可以了；

  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<string.h>
  5 #include<queue>
  6 #include<set>
  7 #include<math.h>
  8 #include<map>
  9 using namespace std;
 10 typedef long long LL;
 11 int ans[100005];
 12 int cnt[100005];
 13 int cnt2[100005];
 14 int tx[100005];
 15 int ask[100005];
 16 int main(void)
 17 {
 18         int T;
 19         scanf("%d",&T);
 20         while(T--)
 21         {
 22                 int n;
 23                 scanf("%d",&n);
 24                 int i,j;
 25                 for(i = 1; i <= n; i++)
 26                 {
 27                         scanf("%d",&ans[i]);
 28                 }
 29                 int cn = 0;
 30                 for(i = 1; i <= sqrt(1.0*n); i++)
 31                 {
 32                         int v = 0;
 33                         if(n%i==0)
 34                         {
 35                                 int k = n/i;
 36                                 for(j = 1; j <= i; j++)
 37                                 {
 38                                         if(!cnt[ans[j]])
 39                                         {
 40                                                 tx[v++] = ans[j];
 41                                         }
 42                                         cnt[ans[j]]++;
 43                                 }
 44                                 bool flag  = false ;
 45                                 int x = i+1;
 46                                 while(true)
 47                                 {
 48                                         for(j = x; j <= i+x-1&& j<=n; j++)
 49                                         {
 50                                                 if(!cnt[ans[j]])
 51                                                 {
 52                                                         flag  = true;
 53                                                         break;
 54                                                 }
 55                                                 cnt2[ans[j]]++;
 56                                         }
 57                                         x = j;
 58                                         for(j = 0; j < v; j++)
 59                                         {
 60                                                 if(cnt[tx[j]]!=cnt2[tx[j]])
 61                                                 {
 62                                                         flag = true;
 63                                                 }
 64                                                 cnt2[tx[j]] = 0;
 65                                         }
 66                                         if(flag || x == n+1)
 67                                         {
 68                                                 break;
 69                                         }
 70                                 }
 71                                 for(j = 0; j  < v; j++)
 72                                 {
 73                                         cnt[tx[j]] = 0;
 74                                 }
 75                                 if(!flag)ask[cn++] = i;
 76                                 if(i!=n/i)
 77                                 {
 78                                         v = 0;
 79                                         for(j = 1; j <= k; j++)
 80                                         {
 81                                                 if(!cnt[ans[j]])
 82                                                 {
 83                                                         tx[v++] = ans[j];
 84                                                 }
 85                                                 cnt[ans[j]]++;
 86                                         }
 87                                         bool flag  = false ;
 88                                         int x = k+1;
 89                                         while(true)
 90                                         {
 91                                                 for(j = x; j <= k+x-1&& j<=n; j++)
 92                                                 {
 93                                                         if(!cnt[ans[j]])
 94                                                         {
 95                                                                 flag  = true;
 96                                                                 break;
 97                                                         }
 98                                                         cnt2[ans[j]]++;
 99                                                 }
100                                                 x = j;
101                                                 for(j = 0; j < v; j++)
102                                                 {
103                                                         if(cnt[tx[j]]!=cnt2[tx[j]])
104                                                         {
105                                                                 flag = true;
106                                                         }
107                                                         cnt2[tx[j]] = 0;
108                                                 }
109                                                 if(flag || x == n+1)
110                                                 {
111                                                         break;
112                                                 }
113                                         }
114                                         for(j = 0; j  < v; j++)
115                                         {
116                                                 cnt[tx[j]] = 0;
117                                         }
118                                         if(!flag)ask[cn++] = k;
119                                 }
120                         }
121                 }
122                 ask[cn++] = n;
123                 sort(ask,ask+cn);
124                 printf("%d",ask[0]);
125                 for(i = 1; i < cn; i++)
126                 {
127                         printf(" %d",ask[i]);
128                 }
129                 printf("
");
130         }
131         return 0;
132 }

油！油！you@

原文地址：https://www.cnblogs.com/zzuli2sjy/p/5927276.html