"Codeforces" E. New Year and Entity enumeration

"topic" E. New Year and Entity enumeration

"Test Instructions" the given set T contains N m long binary numbers, which require a set of S number that contains the set T and satisfies the following conditions: The length <=m, the non-and the result are all in the collection. (Details of the original question)

"Algorithm" Mathematics (Bell number)

The problem of "solution" really does not understand this practice, so it is simple to write.

First, do not consider that s must contain set T.

For a scenario, the bitwise consideration, all digits with the bit I and up to get the smallest number containing that bit, recorded as F[i].

For F[x]≠f[y], there are f[x]&f[y]=0, proof:! (F[x]&f[y]) &f[x] This number contains X-bits and <f[x].

Then the f[x of different bits], either equal or unequal and no overlap, then the number of schemes corresponds to the number of sets divided by 1~m (Bell number).

Bell number: B (n) =σc (n-1,k) *b (k), k=0~n-1. Solve the Complexity O (m^2).

Finally consider t, different bits if the binary number of vertical view is different then its F value must not be the same, so divided into a number of parts of the respective solution and then multiply is the answer.

#include <cstdio>#include<map>#definell Long Longusing namespacestd;Const intmaxn=1010, mod=1e9+7;intc[maxn][maxn],f[maxn],n,m;ll B[maxn];map<ll,int>MP;intMain () {scanf ("%d%d",&m,&N);  for(intI=0; i<n;i++){         for(intj=1; j<=m;j++){            intx; scanf ("%1d",&x); B[J]+ = (1ll*x) <<i; }    }     for(intI=1; i<=m;i++) mp[b[i]]++;  for(intI=0; i<=m;i++) {c[i][0]=1;  for(intj=1; j<=i;j++) c[i][j]= (c[i-1][j-1]+c[i-1][J])%MOD; } f[0]=1;  for(intI=1; i<=m;i++){         for(intj=0; j<i;j++) f[i]= (f[i]+1ll*c[i-1][J]*F[J]%MOD)%MOD; }    intans=1;  for(Map<ll,int>::iterator It=mp.begin (); It!=mp.end (); it++) ans=1ll*ans*f[it->second]%MOD; printf ("%d", ans); return 0;}

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"Codeforces" E. New Year and Entity enumeration