A substring of a string T is defined as:
Given two strings A, B and one integer K, we define S, a set of triples (i, j, k):
You are to give the value of |S| for specific A, B and K.
Input
The input file contains several blocks of data. For each block, the first line contains one integer K, followed by two lines containing strings A and B, respectively. The input file is ended by K=0.
1 ≤ |A|, |B| ≤ 105
1 ≤ K ≤ min{|A|, |B|}
Characters of A and B are all Latin letters.
Output
For each case, output an integer |S|.
Sample Input
2
aababaa
abaabaa
1
xx
xx
0
Sample Output
22 5
对第一个串建立自动机，然后让第二个串在上面跑，记录一下到每个状态时的匹配的长度，
一个状态是很多后缀串的集合，我们只需要当前状态中大于等于K 小于等于匹配长度的串
然后如果fa有长度也大于等于k的串的话，那也可以去
为了避免每次都要向上找fa，打一个lazy标记一波，最后在一遍更新就行了。
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<algorithm>
5 #define ri register int
6 using namespace std;
7 const int N=2e5+5;
8 typedef long long ll;
9 int k,n;
10 char s[N];
11 inline int calc(char c){return c>='a'&&c<='z'?c-'a':c-'A'+26;}
12 struct SAM{
13 int tot,rt,last,len[N],link[N],son[N][60],siz[N],lz[N],cnt[N],rk[N];
14 inline void init(){
15 memset(len,0,sizeof(len)),memset(siz,0,sizeof(siz)),memset(rk,0,sizeof(rk)),memset(son,0,sizeof(son));
16 memset(link,0,sizeof(link)),memset(cnt,0,sizeof(cnt)),memset(lz,0,sizeof(lz)),tot=last=rt=1,len[0]=-1;
17 }
18 inline void expend(int x){
19 int p=last,np=++tot;
20 siz[last=np]=1,len[np]=len[p]+1;
21 while(p&&!son[p][x])son[p][x]=np,p=link[p];
22 if(!p){link[np]=rt;return;}
23 int q=son[p][x],nq;
24 if(len[q]==len[p]+1){link[np]=q;return;}
25 len[nq=++tot]=len[p]+1,memcpy(son[nq],son[q],sizeof(son[q])),link[nq]=link[q];
26 while(p&&son[p][x]==q)son[p][x]=nq,p=link[p];
27 link[q]=link[np]=nq;
28 }
29 inline void topsort(){
30 for(ri i=1;i<=tot;++i)++cnt[len[i]];
31 for(ri i=1;i<=last;++i)cnt[i]+=cnt[i-1];
32 for(ri i=1;i<=tot;++i)rk[cnt[len[i]]--]=i;
33 for(ri i=tot;i;--i)siz[link[rk[i]]]+=siz[rk[i]];
34 }
35 inline void query(){
36 topsort();
37 int p=1,nowlen=0;
38 ll ans=0;
39 for(ri i=1;i<=n;++i){
40 int x=calc(s[i]);
41 if(son[p][x])p=son[p][x],++nowlen;
42 else{
43 while(p&&!son[p][x])p=link[p];
44 if(!p)p=1,nowlen=0;
45 else nowlen=len[p]+1,p=son[p][x];
46 }
47 if(nowlen>=k){
48 ans+=(ll)(nowlen-max(k,len[link[p]]+1)+1)*siz[p];
49 if(len[link[p]]>=k)++lz[link[p]];
50 }
51 }
52 for(ri i=tot;i;--i){
53 p=rk[i];
54 ans+=(ll)lz[p]*siz[p]*(len[p]-max(k,len[link[p]]+1)+1);
55 if(len[link[p]]>=k)lz[link[p]]+=lz[p];
56 }
57 cout<<ans<<'
';
58 }
59 }sam;
60 int main(){
61 while(scanf("%d",&k),k){
62 scanf("%s",s+1),n=strlen(s+1),sam.init();
63 for(ri i=1;i<=n;++i)sam.expend(calc(s[i]));
64 scanf("%s",s+1),n=strlen(s+1),sam.query();
65 }
66 return 0;
67 }