Problem Description
Now you are back,and have a task to do:
Given you a string s consist of lower-case English letters only,denote f(s) as the number of distinct sub-string of s.
And you have some query,each time you should calculate f(s[l...r]), s[l...r] means the sub-string of s start from l end at r.
Given you a string s consist of lower-case English letters only,denote f(s) as the number of distinct sub-string of s.
And you have some query,each time you should calculate f(s[l...r]), s[l...r] means the sub-string of s start from l end at r.
Input
The first line contains integer T(1<=T<=5), denote the number of the test cases.
For each test cases,the first line contains a string s(1 <= length of s <= 2000).
Denote the length of s by n.
The second line contains an integer Q(1 <= Q <= 10000),denote the number of queries.
Then Q lines follows,each lines contains two integer l, r(1 <= l <= r <= n), denote a query.
For each test cases,the first line contains a string s(1 <= length of s <= 2000).
Denote the length of s by n.
The second line contains an integer Q(1 <= Q <= 10000),denote the number of queries.
Then Q lines follows,each lines contains two integer l, r(1 <= l <= r <= n), denote a query.
Output
For each test cases,for each query,print the answer in one line.
Sample Input
2
bbaba
5
3 4
2 2
2 5
2 4
1 4
baaba
5
3 3
3 4
1 4
3 5
5 5
Sample Output
3
1
7
5
8
1
3
8
5
1
Hint
I won't do anything against hash because I am nice.Of course this problem has a solution that don't rely on hash.
Author
WJMZBMR
Source
ps：代码自带大常数=-=
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 #include<algorithm> 5 using namespace std; 6 7 const int N = 2*1e4+10; 8 9 struct QN { 10 int l,r,num; 11 bool operator < (const QN& rhs) const { 12 return l<rhs.l || (l==rhs.l&&r<rhs.r); 13 } 14 }que[N]; 15 char s[N]; 16 int n,sz,last; 17 int fa[N],ch[N][26],l[N],ans[N]; 18 19 void clear() { 20 sz=0; last=++sz; 21 memset(ch,0,sizeof(ch)); 22 memset(l,0,sizeof(l)); 23 } 24 void add(int x) { 25 int c=s[x]-'a'; 26 int p=last,np=++sz; last=np; 27 l[np]=l[p]+1; 28 for(;p&&!ch[p][c];p=fa[p]) ch[p][c]=np; 29 if(!p) fa[np]=1; 30 else { 31 int q=ch[p][c]; 32 if(l[p]+1==l[q]) fa[np]=q; 33 else { 34 int nq=++sz; l[nq]=l[p]+1; 35 memcpy(ch[nq],ch[q],sizeof(ch[q])); 36 fa[nq]=fa[q]; 37 fa[np]=fa[q]=nq; 38 for(;q==ch[p][c];p=fa[p]) ch[p][c]=nq; 39 } 40 } 41 } 42 int read() { 43 char c=getchar(); int x=0; 44 while(!isdigit(c)) c=getchar(); 45 while(isdigit(c)) x=x*10+c-'0',c=getchar(); 46 return x; 47 } 48 int main() { 49 int T,i,j; T=read(); 50 while(T--) { 51 scanf("%s",s); n=read(); 52 for(i=1;i<=n;i++) { 53 que[i].l=read(),que[i].r=read(); 54 que[i].num=i; 55 } 56 sort(que+1,que+n+1); 57 memset(ans,0,sizeof(ans)); 58 j=que[1].l; 59 for(i=1;i<=n;i++) { 60 if(i==1&&que[i].l==que[i-1].l) 61 for(j;j<=que[i].r;j++) add(j-1); 62 else { 63 clear(); 64 for(j=que[i].l;j<=que[i].r;j++) add(j-1); 65 } 66 for(j=1;j<=sz;j++) 67 ans[que[i].num]+=l[j]-l[fa[j]]; 68 } 69 for(int i=1;i<=n;i++) 70 printf("%d ",ans[i]); 71 } 72 return 0; 73 }