BZOJ 1009 GT考试

Description

阿申准备报名参加GT考试，准考证号为N位数X1X2....Xn(0<=Xi<=9),他不希望准考证号上出现不吉利的数字。他的不吉利数学A1A2...Am(0<=Ai<=9)有M位，不出现是指X1X2...Xn中没有恰好一段等于A1A2...Am. A1和X1可以为0

Input

第一行输入N,M,K.接下来一行输入M位的数。 100%数据N<=10^9,M<=20,K<=1000 40%数据N<=1000 10%数据N<=6

Output

阿申想知道不出现不吉利数字的号码有多少种，输出模K取余的结果.

Sample Input

4 3 100
111

Sample Output

81

HINT

 

Source

利用“不吉利数字”构建ac自动机（我好像小题大做了，一个串好像没有必要），构建初始矩阵，接着矩阵乘法即可。

初始矩阵a[i][j]表示走一步从i节点走到j节点的方案数。

code：

 #include<vector>
 #include<queue>
 #include<cstring>
 #include<cstdio>
 #include<cstdlib>
 using namespace std;
 
 #define maxm 25
 char buf[maxm];
 int n,m,rhl,ans;
 
 struct node
 {
     int a[maxm*maxm][maxm*maxm],n,m;
     node() {memset(a,,sizeof(a));n = m = ;}
 
     friend node operator *(node x,node y)
     {
         node z; z.n = x.n; z.m = y.m;
         int i,j,k;
         for (i = ;i <= z.n;++i)
             for (j = ;j <= z.m;++j)
                 for (k = ;k <= x.m;++k)
                     (z.a[i][j] += (long long)x.a[i][k]*(long long)y.a[k][j]%rhl)%=rhl;
         return z;
     }
 
     inline node quick(node x,int k)
     {
         node ret; ret.n = x.n; ret.m = x.m;
         for (int i = ;i <= ret.n;++i) ret.a[i][i] = ;
         for (;k;k>>=,x = x*x)
             if (k & )
                 ret = ret*x;
         return ret;
     }
 
     inline void calc() { for (int i = ;i <= m;++i) (ans += a[][i])%=rhl; }
 }s;
 
 struct trie
 {
     int next[maxm][],fail[maxm],L,root;
     bool end[maxm];
     inline int newnode()
     {
         memset(next[L],-,sizeof(next[L]));
         return ++L-;
     }
 
     inline void init() {L = ; root = newnode();}
 
     inline void insert()
     {
         int len = strlen(buf),now = root,i;
         for (i = ;i < len;++i)
         {
             if (next[now][buf[i]-''] == -) next[now][buf[i]-''] = newnode();
             now = next[now][buf[i]-''];
         }
         end[now] = true;
     }
 
     inline void build()
     {
         int now = root,i; queue <int> team;
         fail[root] = root;
         for (i = ;i < ;++i)
             if (next[now][i] == -) next[now][i] = root;
             else fail[next[now][i]] = root,team.push(next[now][i]);
         while (!team.empty())
         {
             now = team.front(); team.pop();
             for (i = ;i < ;++i)
                 if (next[now][i] == -) next[now][i] = next[fail[now]][i];
                 else fail[next[now][i]] = next[fail[now]][i],team.push(next[now][i]);
         }
     }
 
     inline void ready()
     {
         vector <int> son[maxm]; queue <int> team; int i,now,v,nn;
         for (i = ;i < L;++i) if (fail[i] != i) son[fail[i]].push_back(i);
         team.push(root);
         while (!team.empty())
         {
             now = team.front(); team.pop();
             nn = son[now].size();
             for (i = ;i < nn;++i)
             {
                 v = son[now][i];
                 if (end[now]) end[v] = true;
                 team.push(v);
             }
         }
     }
 
     inline void make()
     {
         int i,j; s.n = s.m = L;
         for (i = ;i < L;++i)
         {
             if (end[i]) continue;
             for (j = ;j < ;++j)
                 if (!end[next[i][j]]) s.a[i+][next[i][j]+]++;
         }
     }
 }ac;
 
 int main()
 {
     freopen("1009.in","r",stdin);
     freopen("1009.out","w",stdout);
     scanf("%d %d %d\n",&n,&m,&rhl);
     ac.init();
     scanf("%s",buf); ac.insert();
     ac.build(); ac.ready(); ac.make();
     s = s.quick(s,n); s.calc();
     printf("%d",ans);
     fclose(stdin); fclose(stdout);
     return ;
 }