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：

  1 #include<vector>
  2 #include<queue>
  3 #include<cstring>
  4 #include<cstdio>
  5 #include<cstdlib>
  6 using namespace std;
  7 
  8 #define maxm 25
  9 char buf[maxm];
 10 int n,m,rhl,ans;
 11 
 12 struct node
 13 {
 14     int a[maxm*maxm][maxm*maxm],n,m;
 15     node() {memset(a,0,sizeof(a));n = m = 0;}
 16 
 17     friend node operator *(node x,node y)
 18     {
 19         node z; z.n = x.n; z.m = y.m;
 20         int i,j,k;
 21         for (i = 1;i <= z.n;++i)
 22             for (j = 1;j <= z.m;++j)
 23                 for (k = 1;k <= x.m;++k)
 24                     (z.a[i][j] += (long long)x.a[i][k]*(long long)y.a[k][j]%rhl)%=rhl;
 25         return z;
 26     }
 27 
 28     inline node quick(node x,int k)
 29     {
 30         node ret; ret.n = x.n; ret.m = x.m;
 31         for (int i = 1;i <= ret.n;++i) ret.a[i][i] = 1;        
 32         for (;k;k>>=1,x = x*x)
 33             if (k & 1)
 34                 ret = ret*x;
 35         return ret;
 36     }
 37 
 38     inline void calc() { for (int i = 0;i <= m;++i) (ans += a[1][i])%=rhl; }
 39 }s;
 40 
 41 struct trie
 42 {
 43     int next[maxm][10],fail[maxm],L,root;
 44     bool end[maxm];
 45     inline int newnode()
 46     {
 47         memset(next[L],-1,sizeof(next[L]));
 48         return ++L-1;
 49     }
 50 
 51     inline void init() {L = 0; root = newnode();}
 52     
 53     inline void insert()
 54     {
 55         int len = strlen(buf),now = root,i;
 56         for (i = 0;i < len;++i)
 57         {
 58             if (next[now][buf[i]-'0'] == -1) next[now][buf[i]-'0'] = newnode();
 59             now = next[now][buf[i]-'0'];
 60         }
 61         end[now] = true;
 62     }
 63 
 64     inline void build()
 65     {
 66         int now = root,i; queue <int> team;
 67         fail[root] = root;
 68         for (i = 0;i < 10;++i)
 69             if (next[now][i] == -1) next[now][i] = root;
 70             else fail[next[now][i]] = root,team.push(next[now][i]);
 71         while (!team.empty())
 72         {
 73             now = team.front(); team.pop();
 74             for (i = 0;i < 10;++i)
 75                 if (next[now][i] == -1) next[now][i] = next[fail[now]][i];
 76                 else fail[next[now][i]] = next[fail[now]][i],team.push(next[now][i]);
 77         }
 78     }
 79 
 80     inline void ready()
 81     {
 82         vector <int> son[maxm]; queue <int> team; int i,now,v,nn;
 83         for (i = 0;i < L;++i) if (fail[i] != i) son[fail[i]].push_back(i);
 84         team.push(root);
 85         while (!team.empty())
 86         {
 87             now = team.front(); team.pop();
 88             nn = son[now].size();
 89             for (i = 0;i < nn;++i)
 90             {
 91                 v = son[now][i];
 92                 if (end[now]) end[v] = true;
 93                 team.push(v);
 94             }    
 95         }
 96     }
 97 
 98     inline void make()
 99     {
100         int i,j; s.n = s.m = L;
101         for (i = 0;i < L;++i)
102         {
103             if (end[i]) continue;
104             for (j = 0;j < 10;++j)
105                 if (!end[next[i][j]]) s.a[i+1][next[i][j]+1]++;
106         }
107     }
108 }ac;
109 
110 int main()
111 {
112     freopen("1009.in","r",stdin);
113     freopen("1009.out","w",stdout);
114     scanf("%d %d %d\n",&n,&m,&rhl);
115     ac.init();
116     scanf("%s",buf); ac.insert();
117     ac.build(); ac.ready(); ac.make();
118     s = s.quick(s,n); s.calc();
119     printf("%d",ans);
120     fclose(stdin); fclose(stdout);
121     return 0;
122 }
View Code

 

posted @ 2015-01-15 13:45  lmxyy  阅读(146)  评论(0)  编辑  收藏  举报