| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 46519 | Accepted: 18533 |
Description
The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. A quote from the book:
Everything seemed normal, but everything asserted itself to be false. Everything seemed normal at first, then the inhuman, maddening loomed up. He would have liked to know where the association that united him with the novel was articulated: on his carpet, constantly assailing his imagination, the intuition of a taboo, the vision of an obscure evil, of a vacant what , of an unsaid: the vision, the avision of an oblivion commanding everything, where reason was abolished: everything seemed normal but...
Perec would probably have scored high (or rather, low) in the following contest. People are asked to write a perhaps even meaningful text on some subject with as few occurrences of a given “word” as possible. Our task is to provide the jury with a program that counts these occurrences, in order to obtain a ranking of the competitors. These competitors often write very long texts with nonsense meaning; a sequence of 500,000 consecutive 'T's is not unusual. And they never use spaces.
So we want to quickly find out how often a word, i.e., a given string, occurs in a text. More formally: given the alphabet {'A', 'B', 'C', …, 'Z'} and two finite strings over that alphabet, a word W and a text T, count the number of occurrences of W in T. All the consecutive characters of W must exactly match consecutive characters of T. Occurrences may overlap.
Input
The first line of the input file contains a single number: the number of test cases to follow. Each test case has the following format:
- One line with the word W, a string over {'A', 'B', 'C', …, 'Z'}, with 1 ≤ |W| ≤ 10,000 (here |W| denotes the length of the string W).
- One line with the text T, a string over {'A', 'B', 'C', …, 'Z'}, with |W| ≤ |T| ≤ 1,000,000.
Output
For every test case in the input file, the output should contain a single number, on a single line: the number of occurrences of the word W in the text T.
Sample Input
3 BAPC BAPC AZA AZAZAZA VERDI AVERDXIVYERDIAN
Sample Output
1 3 0
#include<stdio.h>
#include<string.h>
#include<cmath>
#include<stdlib.h>
#include<time.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<queue>
#include<bitset>
#define ll long long
#define ull unsigned long long
#define qq printf("QAQ\n");
using namespace std;
const int maxn=1e5+5;
const int inf=0x3f3f3f3f;
const ll mod=1e9+7;
const ull hmod1=19260817;
const ull hmod2=19660813;
const double e=exp(1.0);
const double pi=acos(-1);
ull base=133;
ull cbit[maxn],hs[1000005];
char s1[10005],s2[1000005];
ull quick_pow(int n)
{
ull ans=1,a=base;
while(n)
{
if(n&1)ans*=a;
n/=2;
a=a*a;
}
return ans;
}
ull check(int l,int r,ull bit)
{
return (ull)hs[r]-bit*hs[l-1];
// return (ull)hs[r]-(ull)cbit(r-l+1)*hs[l-1];
}
int main()
{
int t;
scanf("%d",&t);
cbit[1]=base;
/*for(int i=2;i<maxn;i++)
cbit[i]=cbit[i-1]*base;*/
while(t--)
{
scanf("%s%s",s1,s2);
ull target=0;
int l1=strlen(s1),l2=strlen(s2);
for(int i=0;i<l1;i++)
target=target*base+(ull)s1[i];
for(int i=0;i<l2;i++)
hs[i]=hs[i-1]*base+s2[i];
int ans=0;
ull bit=quick_pow(l1);
for(int i=0;i+l1<=l2;i++)
// if(check(i,i+l1-1)==target)ans++;
if(check(i,i+l1-1,bit)==target)ans++;
printf("%d\n",ans);
}
return 0;
}
#include<stdio.h>
#include<string.h>
const int maxn=1e6+5;
int n,l,next[maxn];//next 表示 i 位置时的
char a[maxn],b[10005];
void getnext()
{
next[0]=-1;
int k=-1;
for(int i=1;i<n;i++)
{
while(k>-1&&a[k+1]!=a[i])
k=next[k];
if(a[k+1]==a[i])k++;
next[i]=k;
}
}
int kmp()
{
int k=-1,ans=0;
for(int i=0;i<n;i++)
{
while(k>-1&&b[k+1]!=a[i])
k=next[k];
if(b[k+1]==a[i])k++;
if(k==l-1){
ans++;
k=next[k];
}
}
return ans;
}
int main()
{
/*while(scanf("%d",&n)!=EOF)
{
scanf("%s",a);
getnext();
}*/
int t;
scanf("%d",&t);
while(t--)
{
scanf("%s%s",b,a);
n=strlen(a);
l=strlen(b);
getnext();
printf("%d\n",kmp());
}
return 0;
}