[Bzoj5508] Mr toluene string

First, define the status of F [i] [j] represents the length of i strings to j ending number of qualified string found $ F [i] [j] = \ SUM F [i-. 1] [K] $ ( j and k may be adjacent ) , the matrix multiplication can optimize it.

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 #define mod 1000000007
 5 struct ji{
 6     ll a[31][31];
 7 }a,b,c;
 8 ll n,ans;
 9 char s[100001];
10 ji cheng(ji a,ji b){
11     memset(c.a,0,sizeof(c.a));
12     for(int i=0;i<26;i++)
13         for(int j=0;j<26;j++)
14             if (a.a[i][j])
15                 for(int k=0;k<26;k++)c.a[i][k]=(c.a[i][k]+a.a[i][j]*b.a[j][k])%mod;
16     return c;
17 }
18 int main(){
19     scanf("%lld%s",&n,s);
20     for(int i=0;i<26;i++)
21         for(int j=0;j<26;j++)a.a[i][j]=1;
22     for(int i=0;s[i+1];i++)a.a[s[i]-'a'][s[i+1]-'a']=0;
23     for(int i=0;i<26;i++)b.a[i][i]=1;
24     for(n--;n;n>>=1){
25         if (n&1)b=cheng(b,a);
26         a=cheng(a,a);
27     }
28     for(int i=0;i<26;i++)
29         for(int j=0;j<26;j++)ans=(ans+b.a[i][j])%mod;
30     printf("%lld",ans);
31 }

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