(I2
I122
I6
(dp0
Vtable.%l
p1
(Vtable.cpp
p2
S'#include<bits/stdc++.h>\r\nusing namespace std;\r\nint n,ans,newn,k,a[72],b[72],usd[72];\r\nvector<int>A,B,C;\r\nbool used[71];\r\nlong long pw(int x,int stepen)\r\n{\r\n    long long ans=x;\r\n    for(int i=2;i<=stepen;i++)\r\n        ans*=x;\r\n    return ans;\r\n}\r\nset<set<int> >st;\r\nvoid variacii(int i)\r\n{\r\n    if(i==k+1)\r\n    {\r\n        set<int>s;\r\n        for(int i=1;i<=k;i++)\r\n            s.insert(b[i]-1);\r\n        st.insert(s);\r\n        return;\r\n    }\r\n    for(int j=b[i-1]+1;j<=newn;j++)\r\n    {\r\n        if(usd[j]==0)\r\n        {\r\n            usd[j]=1;\r\n            b[i]=j;\r\n            variacii(i+1);\r\n            usd[j]=0;\r\n            b[i]=0;\r\n        }\r\n    }\r\n}\r\nint main()\r\n{\r\n    ios::sync_with_stdio(false);\r\n    cin.tie(NULL);\r\n    cin>>n;\r\n    int a;\r\n    cin>>a;\r\n    for(int i=0;i<a;i++)\r\n    {\r\n        int x;\r\n        cin>>x;\r\n        used[x]=1;\r\n        A.push_back(x);\r\n    }\r\n    cin>>a;\r\n    for(int i=0;i<a;i++)\r\n    {\r\n        int x;\r\n        cin>>x;\r\n        used[x]=2;\r\n        B.push_back(x);\r\n    }\r\n    for(int i=1;i<=2*n;i++)\r\n        if(used[i]==0)\r\n        {\r\n            used[i]=3;\r\n            C.push_back(i);\r\n        }\r\n    sort(A.begin(),A.end());\r\n    sort(B.begin(),B.end());\r\n    k=n-A.size();\r\n    newn=C.size();\r\n    variacii(1);\r\n    for(set<set<int> >::iterator it=st.begin();it!=st.end();it++)\r\n    {\r\n        vector<int>AA,BB;\r\n        AA=A;\r\n        BB=B;\r\n        bool q[71];\r\n        memset(q,0,sizeof(q));\r\n        for(set<int>::iterator itt=(*it).begin();itt!=(*it).end();itt++)\r\n        {\r\n            int now=*itt;\r\n            q[now]=1;\r\n            AA.push_back(C[now]);\r\n        }\r\n        for(int i=0;i<C.size();i++)\r\n            if(q[i]==0)BB.push_back(C[i]);\r\n        sort(AA.begin(),AA.end());\r\n        sort(BB.begin(),BB.end());\r\n        bool l=0;\r\n        for(int i=0;i<n;i++)\r\n            if(AA[i]>BB[i]){l=1;break;}\r\n        if(!l)ans++;\r\n    }\r\n    cout<<ans<<endl;\r\n    return 0;\r\n}\r\n'
p3
tp4
stp5
.