1 #include<algorithm>
2 #include<iostream>
3 #include<cstdlib>
4 #include<cstring>
5 #include<cstdio>
6 #include<string>
7 #include<cmath>
8 #include<ctime>
9 #include<queue>
10 #include<stack>
11 #include<map>
12 #include<set>
13 #define rre(i,r,l) for(int i=(r);i>=(l);i--)
14 #define re(i,l,r) for(int i=(l);i<=(r);i++)
15 #define Clear(a,b) memset(a,b,sizeof(a))
16 #define inout(x) printf("%d",(x))
17 #define douin(x) scanf("%lf",&x)
18 #define strin(x) scanf("%s",(x))
19 #define LLin(x) scanf("%lld",&x)
20 #define op operator
21 #define CSC main
22 typedef unsigned long long ULL;
23 typedef const int cint;
24 typedef long long LL;
25 using namespace std;
26 void inin(int &ret)
27 {
28 ret=0;int f=0;char ch=getchar();
29 while(ch<'0'||ch>'9'){if(ch=='-')f=1;ch=getchar();}
30 while(ch>='0'&&ch<='9')ret*=10,ret+=ch-'0',ch=getchar();
31 ret=f?-ret:ret;
32 }
33 namespace sa//uoj#35
34 {
35 char s[100050];
36 int sa[100050],t[100050],t2[100050],c[100050],n,height[100050];
37 void build_sa(int m)
38 {
39 int *x=t,*y=t2;
40 re(i,0,m-1)c[i]=0;
41 re(i,0,n-1)c[x[i]=s[i]]++;
42 re(i,1,m-1)c[i]+=c[i-1];
43 rre(i,n-1,0)sa[--c[x[i]]]=i;
44 for(int k=1;k<=n;k<<=1)
45 {
46 int p=0;
47 rre(i,n-1,n-k)y[p++]=i;
48 re(i,0,n-1)if(sa[i]>=k)y[p++]=sa[i]-k;
49 re(i,0,m-1)c[i]=0;
50 re(i,0,n-1)c[x[y[i]]]++;
51 re(i,0,m-1)c[i]+=c[i-1];
52 rre(i,n-1,0)sa[--c[x[y[i]]]]=y[i];
53 swap(x,y),x[sa[0]]=0,p=1;
54 re(i,1,n-1)x[sa[i]]=y[sa[i-1]]==y[sa[i]]&&y[sa[i-1]+k]==y[sa[i]+k]?p-1:p++;
55 if(p>=n)break;
56 m=p;
57 }
58 }
59 void build_height()
60 {
61 re(i,0,n-1)c[sa[i]]=i;
62 int k=0;
63 re(i,0,n-1)
64 {
65 if(!c[i])continue;
66 if(k)k--;
67 int j=sa[c[i]-1];
68 while(s[i+k]==s[j+k])k++;
69 height[c[i]]=k;
70 }
71 }
72 void solve()
73 {
74 strin(s);n=strlen(s);
75 build_sa(200);
76 build_height();
77 re(i,0,n-1)printf("%d%c",sa[i]+1,i==n-1?'\n':' ');
78 re(i,1,n-1)printf("%d%c",height[i],i==n-1?'\n':' ');
79 }
80 }
81 namespace acm//bzoj3172
82 {
83 int ch[1000010][26],pre[1000010],last[1000010],sum[1000010],tag[1000010],ed;
84 char s[1000010];
85 int wei[222];
86 void add(int id)
87 {
88 int len=strlen(s),temp=0;
89 re(i,0,len-1)
90 {
91 int c=s[i]-'a';
92 if(!ch[temp][c])ch[temp][c]=++ed;
93 temp=ch[temp][c];
94 sum[temp]++;
95 }
96 wei[id]=temp;
97 tag[temp]=id;
98 }
99 queue<int>h;
100 int sta[1000010],top;
101 void getfail()
102 {
103 re(i,0,25)if(ch[0][i])pre[ch[0][i]]=0,h.push(ch[0][i]);
104 while(!h.empty())
105 {
106 int x=h.front();h.pop();
107 sta[++top]=x;
108 re(i,0,25)
109 {
110 if(!ch[x][i])
111 {
112 ch[x][i]=ch[pre[x]][i];
113 continue;
114 }
115 int vv=ch[x][i];
116 int k=pre[x];
117 pre[vv]=ch[k][i];
118 last[vv]=tag[pre[vv]]?pre[vv]:last[pre[vv]];
119 h.push(vv);
120 }
121 }
122 }
123 int n;
124 void solve()
125 {
126 inin(n);
127 re(i,1,n)strin(s),add(i);
128 getfail();
129 while(top)sum[pre[sta[top]]]+=sum[sta[top]],top--;
130 re(i,1,n)printf("%d\n",sum[wei[i]]);
131 }
132
133 }
134 namespace kmp//bzoj1355
135 {
136 char a[1000010],b[1000010];
137 int n,m,pre[1000010];
138 void getpre(char *s)
139 {
140 int k=0;n=strlen(s+1);
141 re(i,2,n)
142 {
143 while(k&&s[i]!=s[k+1])k=pre[k];
144 if(s[i]==s[k+1])k++;
145 pre[i]=k;
146 }
147 }
148 int kmp(char *chang,char *duan)
149 {
150 getpre(duan);
151 int k=0;n=strlen(chang+1),m=strlen(duan+1);
152 re(i,1,n)
153 {
154 while(k&&chang[i]!=duan[k+1])k=pre[k];
155 if(chang[i]==duan[k+1])
156 {
157 k++;
158 if(k==m)return i-m+1;
159 }
160 }
161 return -1;
162 }
163 void solve()
164 {
165 inin(n);
166 strin(a+1);
167 getpre(a);
168 printf("%d",n-pre[n]);
169 }
170 }
171 namespace exkmp//bzoj3670
172 {
173 char s[1000010],t[1000010];int pre[1000010],Max[1000010],sum[1000010];
174 const int mod=1e9+7;
175 void getpre(char *s)
176 {
177 int len=strlen(s),a=0;
178 pre[0]=len;
179 while(a<len-1&&s[a]==s[a+1])a++;
180 pre[1]=a;
181 a=1;
182 re(k,2,len-1)
183 {
184 int p=a+pre[a]-1,l=pre[k-a];
185 if(k-1+l>=p)
186 {
187 int j=(p-k+1)>0?(p-k+1):0;
188 while(k+j<len&&s[k+j]==s[j])j++;
189 pre[k]=j;
190 a=k;
191 }
192 else pre[k]=l;
193 }
194 }
195 void getextend(char *s,char *t)
196 {
197 int n=strlen(s),m=strlen(t),a=0;
198 getpre(s);
199 while(a<n-1&&s[a]==t[a+1])a++;
200 Max[0]=a;
201 a=1;
202 re(k,1,n-1)
203 {
204 int p=a+Max[a]-1,l=pre[k-a];
205 if(k+l-1>=p)
206 {
207 int j=(p-k+1)>0?(p-k+1):0;
208 while(k+j<m&&s[k+j]==t[j])j++;
209 Max[k]=j;
210 a=k;
211 }
212 else Max[k]=l;
213 }
214 }
215 int n,T;
216 void solve()
217 {
218 inin(n);
219 while(n--)
220 {
221 T++;
222 strin(s);
223 int len=strlen(s);
224 getpre(s);
225 re(i,1,len-1)
226 {
227 int r=min(i<<1,pre[i]+i);
228 if(t[i]!=T)t[i]=T,sum[i]=1;
229 else sum[i]++;
230 if(t[r]!=T)t[r]=T,sum[r]=-1;
231 else sum[r]--;
232 }
233 if(t[0]!=T)t[0]=T,sum[0]=0;
234 LL ans=sum[0]+1;
235 re(i,1,len-1)
236 {
237 if(t[i]!=T)t[i]=T,sum[i]=0;
238 sum[i]+=sum[i-1];ans=ans*(sum[i]+1)%mod;
239 }
240 printf("%lld\n",ans);
241 }
242 }
243 }
244 namespace pam
245 {
246 const int xxxx=600010;//length
247 int ch[xxxx][26];//son
248 int pre[xxxx];//fail
249 int sum[xxxx];//times
250 int len[xxxx];//Max_length
251 int ss[xxxx];//string
252 int num[xxxx];//numbers
253 int last,n,ed=-1;
254 int newnode(int x)//you_know
255 {
256 len[++ed]=x;
257 return ed;
258 }
259 void init()//pretreatment
260 {
261 newnode(0),newnode(-1);
262 last=n=0;ss[n]=-1,pre[0]=pre[1]=1;
263 }
264 int getfail(int x)//you_know
265 {
266 while(ss[n-len[x]-1]!=ss[n])x=pre[x];
267 return x;
268 }
269 void add(int c)//add_a_char
270 {
271 c-='a';
272 ss[++n]=c;
273 int temp=getfail(last);
274 if(!ch[temp][c])
275 {
276 int now=newnode(len[temp]+2);
277 int k=getfail(pre[temp]);
278 pre[now]=ch[k][c];
279 ch[temp][c]=now;
280 num[now]=num[pre[now]]+1;
281 }
282 last=ch[temp][c];
283 sum[last]++;
284 }
285 void add(char *s)//add_a_string
286 {
287 int limit=strlen(s);
288 re(i,0,limit-1)add(s[i]);
289 }
290 void count()//clac_times
291 {
292 rre(i,ed,1)sum[pre[i]]+=sum[i];
293 }
294 }
295 namespace manacher
296 {
297 const int xxxx=5000050;
298 char s[xxxx];
299 int n,ans,len[xxxx];
300 void solve()
301 {
302 strin(s+1);int m=strlen(s+1);
303 rre(i,m,1)s[(i<<1)-1]=s[i],s[i<<1]='$';
304 s[0]='$';n=m<<1;
305 for(int i=1,j=0,k;i<n;i+=k,j=max(j-k,0))
306 {
307 while(i+j+1<=n&&s[i-j-1]==s[i+j+1])j++;
308 len[i]=j;ans=max(ans,len[i]);
309 for(k=1;k<=j&&len[i]-k!=len[i-k];k++)
310 len[i+k]=min(len[i]-k,len[i-k]);
311 }
312 printf("%d",ans);
313 }
314 }
315 int main()
316 {
317 return 0;
318 }
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