题意，，，给你 M 个字符去组成密码，密码的长度规定为 N，所组成的密码要求至少 含有 M 个字符里面的 K 个，字符可能包含，不能重复计算；

这个    有点难度；因为  M 很小，所以，可以用二进制来做，当然  内存可能会爆，如果不注意好的话； 上题是要不包含字符，而这题是要求包含字符，那么对应的每一个状态是不是又多了一维变量，每个状态包含了多少个字符，从 1～k --》

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<algorithm>
#define mod 20090717
using namespace std;struct date
{date *next[26],*fail;int in,cnt;
}tree[1221],*que[1221],*root;
int N,M,P,tail,head,total;
char dir[20][20];date *creat_node( )
{for( int i = 0; i < 26; i++ )tree[total].next[i] = NULL;tree[total].fail    = NULL;tree[total].in = total;tree[total].cnt = 0;return &tree[total++];
}void inint( )
{tail = head = total = 0;root = creat_node();root->fail = root;
}void insert( char *word,int i )
{date *temp = root;while( *word ){int k = *word - 'a';if( temp->next[k] == NULL )temp->next[k] = creat_node();temp = temp->next[k];word++;}temp->cnt |= (1<<i);
}void build_AC( )
{que[tail++] = root;while( tail > head ){date *temp = que[head++];for( int i = 0; i < 26; i++ )if( temp->next[i] ){if( temp == root ) temp->next[i]->fail = root;else               temp->next[i]->fail = temp->fail->next[i];temp->next[i]->cnt = temp->next[i]->cnt|temp->fail->next[i]->cnt|temp->cnt;que[tail++] = temp->next[i];}else {if( temp == root ) temp->next[i] = root;else               temp->next[i] = temp->fail->next[i];}}
}int map[1500];
void work( )
{for( int i = 0; i <= 1500; i++ )map[i] = map[i>>1] + (i&1);
}int  dp[27][121][1500];
void DP( )
{int i,j,k,l,pos,res;pos = (1<<M );for (i = 0; i <= N; ++i)  for (j = 0; j < total; ++j)  for (k = 0; k < (1<<M); ++k)  dp[i][j][k] = 0;  dp[0][0][0] = 1;for( i = 1; i <= N;   i++ )for( j = 0; j <total; j++ )for( k = 0; k < pos;  k++ ){if( dp[i-1][j][k] == 0 ) continue;int a = dp[i-1][j][k];for( l = 0; l <  26;  l++ ){int b = tree[j].next[l]->in;int c = tree[j].next[l]->cnt;dp[i][b][k|c] = ( a + dp[i][b][k|c] )%mod;}}res = 0; for( i = 0; i <  total; i++ )for( j = 0; j <  pos; j++ )if( map[j] >= P )res = ( res + dp[N][i][j] )%mod;printf("%d\n",res);
}int main( )
{work();while( scanf("%d%d%d",&N,&M,&P) != EOF ){if( N+M+P == 0 )break;inint();for( int i = 0; i < M; i++ )scanf("%s",&dir[i]),insert( dir[i],i );build_AC( );DP( );}return 0;
}