HDU 5908 Abelian Period (暴力)

题意：给定对于一个数字串S和一个正整数k，如果S可以分成若干个长度为k的连续子串，且这些子串两两匹配，那么我们称k是串S的一个完全阿贝尔周期。

给定一个数字串S，请找出它所有的完全阿贝尔周期。匹配就是含有相同的数字。

析：枚举k，首先k必须是 n 的约数，然后就能算出每个数字应该出现多少次。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const LL mod = 10000000000007;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int a[maxn];
bool vis[maxn];

bool judge(int x){
    map<int, int> mp1, mp2;
    for(int i = 0; i < x; ++i)  ++mp1[a[i]];

    for(int i = x; i < n; i += x){
        mp2.clear();
        for(int j = i; j < i+x; ++j) ++mp2[a[j]];
        if(mp1 != mp2)  return false;
    }
    return true;
}

int main(){
    int T;  cin >> T;
    while(T--){
        scanf("%d", &n);
        for(int i = 0; i < n; ++i)  scanf("%d", a+i);

        memset(vis, false, sizeof vis);
        vector<int> v;
        for(int i = 1; i <= n/2; ++i){
            if(!vis[i] && n % i == 0)  if(judge(i)){
                vis[i] = true;
                for(int j = i+i; j <= n/2; j += i)  if(n % j == 0){
                    vis[j] = true;
                }
            }
        }

        vis[n] = true;
        bool ok = false;
        for(int i = 1; i <= n; ++i){
            if(vis[i] && ok)  printf(" %d", i);
            else if(vis[i]){  printf("%d", i);  ok = true; }
        }
        printf("
");
    }
    return 0;
}