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,0,sizeof(a));n = m = 0;}

friend node operator *(node x,node y)
{
node z; z.n = x.n; z.m = y.m;
int i,j,k;
for (i = 1;i <= z.n;++i)
for (j = 1;j <= z.m;++j)
for (k = 1;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 = 1;i <= ret.n;++i) ret.a[i][i] = 1;
for (;k;k>>=1,x = x*x)
if (k & 1)
ret = ret*x;
return ret;
}

inline void calc() { for (int i = 0;i <= m;++i) (ans += a[1][i])%=rhl; }
}s;

struct trie
{
int next[maxm][10],fail[maxm],L,root;
bool end[maxm];
inline int newnode()
{
memset(next[L],-1,sizeof(next[L]));
return ++L-1;
}

inline void init() {L = 0; root = newnode();}

inline void insert()
{
int len = strlen(buf),now = root,i;
for (i = 0;i < len;++i)
{
if (next[now][buf[i]-'0'] == -1) next[now][buf[i]-'0'] = newnode();
now = next[now][buf[i]-'0'];
}
end[now] = true;
}

inline void build()
{
int now = root,i; queue <int> team;
fail[root] = root;
for (i = 0;i < 10;++i)
if (next[now][i] == -1) 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 = 0;i < 10;++i)
if (next[now][i] == -1) 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 = 0;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 = 0;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 = 0;i < L;++i)
{
if (end[i]) continue;
for (j = 0;j < 10;++j)
if (!end[next[i][j]]) s.a[i+1][next[i][j]+1]++;
}
}
}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 0;
}

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