题目大意

对于任意长度n的小写字母字符串，不包含给定字符串个数。

解题思路

f[i][j]表示第i个字符，与给定串匹配了j位的方案数。预先用kmp预处理好填一个字符会匹配到哪里，dp转移即可。

code

#include<cstdio>
#include<cstring>
#include<algorithm>
#define min(a,b) ((a<b)?a:b)
#define fo(i,j,k) for(int i=j;i<=k;i++)
#define fd(i,j,k) for(int i=j;i>=k;i--)
using namespace std;
int const maxn=1e4,maxl=100;
int n,m,f[maxn+10][maxl+10],fail[maxl+10][26+10],mod=1e9+7;
char s[maxl+10];
int main(){
    freopen("helloworld.in","r",stdin);
    freopen("helloworld.out","w",stdout);
    for(;scanf("%d",&n)!=EOF;){
        scanf("%s",s+1);m=strlen(s+1);
        int ok=1;
        fo(i,1,m)if((s[i]<'a')||(s[i]>'z')){ok=0;break;}
        if(!ok){
            long long ans=1;
            fo(i,1,n)ans=(ans*26)%mod;
            printf("%lld\n",ans);
            continue;
        }
        fo(j,0,25)fail[0][j]=fail[1][j]=0;
        fo(i,2,m)
            fo(j,0,25){
                fail[i][j]=fail[i-1][s[i-1]-'a'];
                for(;fail[i][j]&&(s[fail[i][j]+1]!='a'+j);fail[i][j]=fail[fail[i][j]][s[fail[i][j]]-'a']);
                if(s[fail[i][j]+1]=='a'+j)fail[i][j]++;
            }
        fo(i,0,n)fo(j,0,m)f[i][j]=0;
        f[0][0]=1;
        fo(i,0,n-1)
            fo(j,0,m-1)
                if(f[i][j])fo(k,0,25){
                    if(k==s[j+1]-'a'){
                        if(j+1!=m)f[i+1][j+1]=(f[i+1][j+1]+f[i][j])%mod;
                    }else f[i+1][fail[j+1][k]]=(f[i+1][fail[j+1][k]]+f[i][j])%mod;
                }
        int ans=0;
        fo(j,0,m-1)ans=(ans+f[n][j])%mod;
        printf("%d\n",ans);
    }
    return 0;
}