hdu 5908 Abelian Period(暴力 + map优化)

Abelian Period

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

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

把一个长为n的数组切成i段，这i段都要满足以下条件：
1.每一段的长度相同；
2.每一段中的某个数字的出现次数要两两相同；
求所有可能的切割方案，并输出每个方案的n / i.

题目本身不难，思路也很耿直，枚举k(1~n的整数)，若是n的约数就直接把数组平均切成k段，判断每一段各个数字的出现次数是否相等即可，理论上时间复杂度可以过。

然而一开始记录是开的数组，由于memset的存在，时间死活压不下来，最后用map才搞定（事实上使用map也是这道题卡时间的关键）。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <map>#include <cmath>#define MAX 100010#define M 100005#define INF 0x3f3f3f3f#define PI acos(-1.0)#define eps 1e-8using namespace std;int arr[M];map<int, int> cmp, tmp;int main(){    int T, n;    scanf("%d", &T);    while(T--)    {        scanf("%d", &n);        for(int i = 0; i < n; i++)            scanf("%d", &arr[i]);        int i = n, a, k, j;        for(; i > 0; i--)        {            if(n % i)                continue;            cmp.clear();            k = n / i;            for(j = 0; j < k; j++)            {                cmp[arr[j]]++;            }            tmp = cmp, a = 0;            for(; j < n - k + 1; j += k)            {                tmp = cmp;                for(a = j; a < j + k; a++)                {                    if(--tmp[arr[a]] < 0)//如果中间有的数字多出来就跳出循环                        break;                }                if(tmp[arr[a]] < 0)//同上                    break;            }            if(tmp[arr[a]] && tmp[arr[a]] < 0)//同上，进行下一循环                continue;            printf("%d", k);            i--;            break;        }        for(; i > 0; i--)//只是为了保证输出格式写成这样，上下两段循环里的内容是一样的。        {            if(n % i)                continue;            k = n / i;            cmp.clear();            for(j = 0; j < k; j++)            {                cmp[arr[j]]++;            }            tmp = cmp, a = 0;            for(; j < n - k + 1; j += k)            {                tmp = cmp;                for(a = j; a < j + k; a++)                {                    if(--tmp[arr[a]] < 0)                        break;                }                if(tmp[arr[a]] < 0)                    break;            }            if(tmp[arr[a]] && tmp[arr[a]] < 0)                continue;            printf(" %d", k);        }printf("\n");    }    return 0;}

运行结果：