2016/10/20

hdu5908

划分子序列为等长子串，这些子串要求两两匹配，匹配的含义是各个数字出现的次数一样。10w规模。

显然剪枝的方法就是利用子长度为总长的约数。匹配其实所有都和第一个子序列去匹配就好了。先++统计第一个，其余的

遇到一次--，全部是0就匹配成功。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cstring>
 5 #include<string>
 6 #include<cmath>
 7 #include<set>
 8 #include<map>
 9 #include<stack>
10 #include<queue>
11 #include<algorithm>
12 #include<sstream>
13 using namespace std;
14 #define ll long long
15 #define LL long long
16 #define inf 0x3f3f3f3f
17 #define llinf 0x3f3f3f3f3f3f3f3f
18 #define FOR(i,a,b) for(int i=a;i<=b;i++)
19 #define FORD(i,a,b) for(int i=b;i>=a;i--)
20 #define fp freopen("/Volumes/未命名2/Downloads/acm/in.txt","r",stdin)
21 #define ptarr(a,x,y) for(int _=x;_<=y;_++) if(_!=y) cout<<a[_]<<" ";else cout<<a[_]<<endl;
22 #define pt1(a) cout<<a<<endl;
23 #define pt2(a,b) cout<<a<<" "<<b<<endl;
24 #define pt3(a,b,c) cout<<a<<" "<<b<<" "<<c<<endl;
25 #define pt4(a,b,c,d) cout<<a<<" "<<b<<" "<<c<<" "<<d<<endl;
26 #define ln1 cout<<"----------------------"<<endl;
27 #define ln2 cout<<"~~~~~~~~~~~~~~~~~~~~~~"<<endl;
28 #define CLR(a,b) memset(a,b,sizeof(a));
29 struct node{
30     int a, b;
31     bool operator<(const node& rhs)const {
32         if(a==rhs.a)
33             return b<rhs.b;
34         else return a<rhs.a;
35     }
36 };
37 int T,n,m,K;
38 #define maxn 200000
39 int a[maxn],cnt[maxn];
40 void read(){
41     cin>>n;
42     FOR(i,1,n) cin>>a[i];
43 }
44 void solve(){
45     vector<int>ans;
46     FOR(len,1,n)
47     {
48         if(n%len!=0) continue;
49         else
50         {
51             int l=1,r=len,flg=1;
52             for(l=1,r=len;r<=n;l+=len,r+=len)
53             {
54                 FOR(k,1,len) cnt[a[k]]++;
55                 FOR(k,l,r) if(cnt[a[k]]) cnt[a[k]]--;
56                 FOR(k,1,len) if(cnt[a[k]]!=0) cnt[a[k]]=0,flg=0;
57
58             }
59             if(flg) ans.push_back(len);
60             FOR(k,1,len) cnt[a[k]]=0;
61         }
62     }
63     int sz=ans.size();
64     FOR(k,0,sz-1)
65     {
66         if(k!=sz-1) printf("%d ",ans[k]);
67         else printf("%d\n",ans[k]);
68     }
69 }
70 int main()
71 {
72     cin>>T;
73     while(T--)
74     {
75         read();
76         solve();
77     }
78     return 0;
79 }