But small can the school library administrator, and now he inherited a very difficult task. Since the schools need some materials, the principal need in the article inspection
search some information. Chancellor total cocoa N to a small article, each article as a string. Now, the president needs him to find such a word, it is at least
M N articles in this article appeared in the article, and word length L. However, the workload is very large, but the president has little cocoa urgent need to complete this task
. Now he turned to you, you need to write a program to complete this arduous task
Input
row 13 positive integers N, M, L, represents the number of articles, the word appears at least in length and M articles in each word.
Next N lines of a string that represents the article.
1≤N, M≤2000, L≤1000. The length of each article is no greater than 1000, are lowercase letters
Output
only one line, the number of words expressed satisfying the search condition.
The Input the Sample
3 2 2
NOIP
istudycpp
imacppstudent
the Sample the Output
5
// These five words are: st, tu, ud, pp , cp.
#include <the iostream>
#include <stdio.h>
#include <string.h>
the using namespace STD;
typedef Long Long LL;
const int MaxN = 2005;
const int MaxM = 1000005;
const int MINMOD = 999 973;
const = LL MaxMod 499999999999993ll;
struct the Node
{
int NO, cnt; // NO is where the word article number, cnt is the number of occurrences of the word
LL s; // s is a word hashh value
int next; // resolve the conflict with the open hash
} h [ MaxM];
int N, M, L, TOT, ANS;
int first [MaxM];
char STR [MaxN] [MaxN];
void the Add (U int, int x, LL S)
// in the article in the x short hashh value appears u, s word length hashh value added hashh table
{
++ TOT;
H [TOT] .s = s;
H [TOT] = .cnt. 1;
H [TOT] .no = X;
h [tot] .next = first [ u];
First [U] = TOT;
}
void hashh (X int, U int, LL S)
{int I;
for (I = First [U]; I = 0;! I = H [I] .next)
IF (S == h [i] .s) // appear in hashh table over
{
IF (h [i] == x .no) return;
// x also appears in the first article, do not have control
h [i] X = .no;
H [I] .cnt ++; // number appears plus. 1
return;
}
the add (U, X, S);
}
void the init ()
{
int I;
Scanf ( "% D% D% D" , & N, & M, & L);
for (I =. 1; I <= N; I ++)
Scanf ( "% S", & STR [I]);
ANS = 0;
}
void Work ()
{int I, J, U , K, K1;
LL S, K2;
K1 = K2 =. 1;
for (J = 0; J <-L. 1; J ++)
{
K1 = K1 * 26 is MINMOD%; // magnitude
k2 = k2 * 26% MaxMod;
}
for(i=1;i<=N;i++)
{
u=s=0;
for (j=0;j<L;j++)
{
u=(u*26+str[i][j])%MinMod;//短hashh值
s=(s*26+str[i][j])%MaxMod;//长hashh值
}
hashh(i,u,s);
k=strlen(str[i]);
for(j=L;j<k;j++)
{
u=((u-str[i][j-L]*k1)%MinMod+MinMod)%MinMod;//前面去一位后面加一位
s=((s-str[i][j-L]*k2)%MaxMod+MaxMod)%MaxMod;
u=(u*26+str[i][j])%MinMod;
s=(s*26+str[i][j])%MaxMod;
hashh(i,u,s);
}
}
}
void output()
{
int i;
for(i=1;i<=tot;i++)
if(h[i].cnt>=M) ++ans;
printf("%d\n",ans);
}
int main(void)
{
init();
work();
output();
return 0;
}
#include<bits/stdc++.h>
#define ll long long
#define mod 28282789
#define L 1001
using
namespace
std;
inline
void
read(
int
&x){
int
datta=0;
char
chchc=
getchar
();
bool
okoko=0;
while
(chchc<
'0'
||chchc>
'9'
){
if
(chchc==
'-'
)okoko=1;chchc=
getchar
();}
while
(chchc>=
'0'
&&chchc<=
'9'
){datta=datta*10+chchc-
'0'
;chchc=
getchar
();}
x=okoko?-datta:datta;
}
int
n,m,l,len,times[mod+10],ans;
int
clear[L],nc;
ll base=26,q[L],hx[L];
char
c[L];
bool
did[mod+10];
ll fk(
int
a,
int
b){
return
(hx[b]-(hx[a-1]*q[b-a+1])%mod+mod)%mod;}
int
main(){
read(n),read(m),read(l);q[0]=1;
for
(
int
i=1;i<=L;i++)q[i]=(q[i-1]*base)%mod;
for
(
int
i=1;i<=n;i++){
scanf
(
"%s"
,c+1);
len=
strlen
(c+1);
for
(
int
i=1;i<=len;i++)hx[i]=(hx[i-1]*base+c[i]-
'a'
+1)%mod;
for
(
int
i=1;i+l-1<=len;i++){
ll tp=fk(i,i+l-1);
if
(!did[tp]){
did[tp]=
true
;clear[++nc]=tp;
if
(++times[tp]==m)ans++;
}
}
for
(
int
i=1;i<=nc;i++){did[clear[i]]=
false
;}
nc=0;
}
printf
(
"%d\n"
,ans);
return
0;
}