Bzoj3530 [SDOI2014] Count

Bzoj3530 [SDOI2014] Count

 

3530: [Sdoi2014] Count Description

We call a positive integer N as the lucky number. If and only when its decimal representation does not contain any element in the number string set S, it serves as its sub-string. For example, when S = (22,333,023 3), 233 is the lucky number. 2333, 20233, and 3223 are not the lucky number.
Given N and S, calculate the number of lucky winners not greater than N.

Input

The first line of the input contains an integer N.
The next row contains an integer M, indicating the number of elements in S.
In the next M row, each row has a number string, indicating an element in S.

Output

Output an integer in a row, indicating the value of the modulo 109 + 7.

Sample Input20
3
2
3
14 Sample Output14HINT

In the following table, l represents the length of N, and L represents the sum of the lengths of all strings in S.

1 <= l <= 1200, 1 <= M <= 100, 1 <= L <= 1500

Source

Round 1 day 1

AC automation + Dynamic Planning, great ideas

First, add the mismatched edge to the Trie tree to form a new graph.

Then, dynamic planning is performed in two cases, namely, the number of digits is less than l and the number is equal to l. When the number of digits is less than l, there is no limit, and it can be transferred directly. When the number of digits is equal to l, you need to add one dimension to indicate whether the number of first I bits is equal to the first I bits of n. (For details about transfer, refer to the Code)

Note that the lucky number does not have a leading zero, so the first step in the figure cannot go through t [1] [0].

#include
 
  #include
  
   #include
   
    #include
    
     #include
     
      #include#include
      
       #define F(i,j,n) for(int i=j;i<=n;i++)#define D(i,j,n) for(int i=j;i>=n;i--)#define ll long long#define pa pair
       
        #define maxn 1510#define mod 1000000007using namespace std;int t[maxn][10],go[maxn],f[1210][maxn][2],a[maxn];int tot=1,n,l,ans=0;char s[maxn];bool v[maxn];queue
        
          q;inline void insert(){scanf("%s",s);int len=strlen(s),now=1;F(i,0,len-1){int x=s[i]-'0';if (!t[now][x]) t[now][x]=++tot;now=t[now][x];}v[now]=1;}inline void bfs(){q.push(1);while (!q.empty()){int x=q.front(),y,j;q.pop();v[x]|=v[go[x]];F(i,0,9){j=go[x];while (j&&!t[j][i]) j=go[j];if (t[x][i]){go[y=t[x][i]]=j?t[j][i]:1;q.push(y);}else t[x][i]=j?t[j][i]:1;}}}int main(){scanf("%s",s);l=strlen(s);F(i,0,l-1) a[i]=s[i]-'0';scanf("%d",&n);F(i,1,n) insert();bfs();memset(f,0,sizeof(f));F(i,1,9) if (!v[t[1][i]]) f[1][t[1][i]][0]+=1;F(i,1,l-2) F(j,1,tot) F(x,0,9) if (!v[t[j][x]])(f[i+1][t[j][x]][0]+=f[i][j][0])%=mod;F(i,1,l-1) F(j,1,tot) (ans+=f[i][j][0])%=mod;memset(f,0,sizeof(f));F(i,1,a[0]) if (!v[t[1][i]]) f[1][t[1][i]][i==a[0]]+=1;F(i,1,l-1) F(j,1,tot) F(x,0,9) if (!v[t[j][x]]){(f[i+1][t[j][x]][0]+=f[i][j][0])%=mod;if (x