ZOJ 3494 BCD Code (AC自动机+数位DP，5级)

BCD Code

---

Time Limit: 5 Seconds      Memory Limit: 65536 KB

---

Binary-coded decimal (BCD) is an encoding for decimal numbers in which each digit is represented by its own binary sequence. To encode a decimal number using the common BCD encoding, each decimal digit is stored in a 4-bit nibble:

Decimal:    0     1     2     3     4     5     6     7     8     9BCD:     0000  0001  0010  0011  0100  0101  0110  0111  1000  1001

Thus, the BCD encoding for the number 127 would be:

 0001 0010 0111

We are going to transfer all the integers from A to B, both inclusive, with BCD codes. But we find that some continuous bits, named forbidden code, may lead to errors. If the encoding of some integer contains these forbidden codes, the integer can not be transferred correctly. Now we need your help to calculate how many integers can be transferred correctly.

Input

There are multiple test cases. The first line of input is an integer T ≈ 100 indicating the number of test cases.

The first line of each test case contains one integer N, the number of forbidden codes ( 0 ≤N ≤ 100). Then N lines follow, each of which contains a 0-1 string whose length is no more than 20. The next line contains two positive integersA and B. Neither A or B contains leading zeros and 0 <A ≤ B < 10²⁰⁰.

Output

For each test case, output the number of integers between A and B whose codes do not contain any of theN forbidden codes in their BCD codes. For the result may be very large, you just need to output itmod 1000000009.

Sample Input

31001 101001 100111111 100

Sample Output

3998

References

• http://en.wikipedia.org/wiki/Binary-coded_decimal

---

Author: GUAN, Yao

思路：先用AC自动机记录fobiden 的BCD码，然后预处理出10进制的BCD转移态，

           然后一个简单的数位DP就搞定了。

#include<iostream>#include<cstring>#include<cstdio>#include<queue>#define FOR(i,a,b) for(int i=a;i<=b;++i)#define clr(f,z) memset(f,z,sizeof(f))#define LL long longusing namespace std;const int mm=2013;const int MOD=1000000009;class AC{public://f 失败指针  int ch[mm][2],f[mm];  bool val[mm];  int sz;  AC()  {    sz=1;clr(ch[0],0);val[0]=0;  }  void clear()  {    sz=1;clr(ch[0],0);val[0]=0;  }  int idx(char x)  {    return x-'0';  }  void insert(char*s)  {    int u=0,c;    for(int i=0;s[i];++i)    {      c=idx(s[i]);      if(!ch[u][c])//add a node      { val[sz]=0;        clr(ch[sz],0);ch[u][c]=sz++;      }      u=ch[u][c];    }    val[u]=1;  }  void getFail()  { int u=0,v,r;    queue<int>Q;    f[0]=0;    FOR(i,0,1)    if(ch[0][i])Q.push(ch[0][i]),f[ ch[0][i] ]=0;    while(!Q.empty())    {      r=Q.front();Q.pop();      val[r]|=val[ f[r] ];//其子集符合即可      FOR(c,0,1)      {        u=ch[r][c];        if(!u){ ch[r][c]=ch[ f[r] ][c]; continue;}        Q.push(u);        v=f[r];        while(v&&!ch[v][c])v=f[v];//从父节点找存在的FAIl        f[u]=ch[v][c];      }    }  }  int bcd[mm][10];  int chage(int fa,int num)  {    if(val[fa])return -1;    int cur=fa;    for(int i=3;i>=0;--i)    {      if(val[ ch[cur][(num>>i)&1] ])return -1;      cur=ch[cur][ (num>>i)&1 ];    }    return cur;  }  void get_BCD()///从某节点开始的转移态  {    FOR(i,0,sz-1)    FOR(j,0,9)    bcd[i][j]=chage(i,j);  }  void BCD_out()  {    FOR(i,0,sz-1)    FOR(j,0,9)    printf("bcd=%d %d %d \n",i,j,bcd[i][j]);  }  LL dp[210][mm];int bit[210],pos;  LL DP(int pp,int s,bool big,bool nozero)  {    if(pp == 0)return 1;    if(big&&dp[pp][s]!=-1)return dp[pp][s];    LL ans = 0;    if(nozero)    {      if(bcd[s][0]!=-1)ans += DP(pp-1,bcd[s][0],big || bit[pp]!=0,nozero);        ans %= MOD;    }    else    {        ans += DP(pp-1,s,big || bit[pp]!=0,nozero);        ans %= MOD;    }    int kn= big?9:bit[pp];    for(int i = 1;i<=kn;i++)    {        if(bcd[s][i]!=-1)        {            ans += DP(pp-1,bcd[s][i],big||i!=kn,1);            ans %=MOD;        }    }    if(big&&nozero)dp[pp][s] = ans;///important big is not mean nozero    return ans;  }  LL get(char*s)  { clr(dp,-1);    pos=0;    int len=strlen(s);    for(int i=len-1;i>=0;--i)    bit[++pos]=s[i]-'0';    //return dfs(pos,0,1,1);    return DP(pos,0,0,0);  }}ac;char s[222];int main(){  int cas,n;  while(~scanf("%d",&cas))  {    FOR(ca,1,cas)    { ac.clear();      scanf("%d",&n);      FOR(i,1,n)      {        scanf("%s",s);        ac.insert(s);      }      ac.getFail();      ac.get_BCD();      //ac.BCD_out();      LL ans=0;      scanf("%s",s);      int len=strlen(s);      for(int i=len-1;i>=0;--i)        if(s[i]>'0'){--s[i];break;}        else s[i]='9';      //cout<<s<<endl;      ans-=ac.get(s);      scanf("%s",s);      ans+=ac.get(s);      ans%=MOD;      if(ans<0)ans+=MOD;      printf("%lld\n",ans);    }  }}