bzoj 3530: [Sdoi 2014] 개수

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 #define M 1509
 5 #define MO 1000000007
 6 #define ll long long
 7 using namespace std;
 8 char a[M];
 9 int b[M],c[M],n,shu[M][10],tot=1,m,n1,fail[M],q[M],f[1202][1505][3];
10 ll ans;
11 bool bo[M];
12 void dp(int a1)
13 {
14     for(int i=1;i<=tot;i++)
15       for(int l=0;l<=1;l++)
16        {
17         if(bo[i]||!f[a1-1][i][l])
18           continue;
19         for(int j=0;j<=9;j++)
20           {
21             int k=i;
22             for(;!shu[k][j];k=fail[k]);
23               f[a1][shu[k][j]][l+j>b[a1]]=(f[a1][shu[k][j]][l+j>b[a1]]+f[a1-1][i][l])%MO;
24             if(!j)
25               f[a1][shu[k][j]][2]=(f[a1][shu[k][j]][2]+f[a1-1][i][l])%MO;
26           }
27        }
28     return;
29 }
30 int main()
31 {
32     scanf("%s",a+1);
33     n=strlen(a+1);
34     for(int i=1;i<=n;i++)
35       b[n-i+1]=a[i]-'0';
36     scanf("%d",&m);
37     for(int i=1;i<=m;i++)
38       {
39         scanf("%s",a+1);
40         n1=strlen(a+1);
41         for(int j=1;j<=n1;j++)
42           c[n1-j+1]=a[j]-'0';
43         int now=1;
44         for(int j=1;j<=n1;j++)
45           if(shu[now][c[j]])
46             now=shu[now][c[j]];
47           else
48             {
49                 shu[now][c[j]]=++tot;
50                 now=tot;
51             }
52         bo[now]=1;
53       }
54     for(int i=0;i<=9;i++)
55       shu[0][i]=1;
56     int h=0,t=1;
57     q[1]=1;
58     for(;h<t;)
59       {
60         h++;
61         int p=q[h];
62         for(int i=0;i<=9;i++)
63           if(shu[p][i])
64             {
65                 q[++t]=shu[p][i];
66                 int k=fail[p];
67                 for(;!shu[k][i];k=fail[k]);
68                 fail[shu[p][i]]=shu[k][i];
69             }
70       }
71     f[0][1][0]=1;
72     for(int i=1;i<=n;i++)
73       dp(i);
74     for(int i=1;i<n;i++)
75       for(int j=1;j<=tot;j++)
76         if(!bo[j])
77           ans=(ans+(ll)f[i][j][0]+f[i][j][1]-f[i][j][2])%MO;
78     for(int i=1;i<=tot;i++)
79       if(!bo[i])
80         ans=(ans+f[n][i][0]-f[n][i][2])%MO;
81     printf("%lld
",ans);
82     return 0;
83 }