HDU1686Oulipo
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Oulipo
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5535 Accepted Submission(s): 2211
Tout avait Pair normal, mais tout s’affirmait faux. Tout avait Fair normal, d’abord, puis surgissait l’inhumain, l’affolant. Il aurait voulu savoir où s’articulait l’association qui l’unissait au roman : stir son tapis, assaillant à tout instant son imagination, l’intuition d’un tabou, la vision d’un mal obscur, d’un quoi vacant, d’un non-dit : la vision, l’avision d’un oubli commandant tout, où s’abolissait la raison : tout avait l’air normal mais…
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.
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.
3BAPCBAPCAZAAZAZAZAVERDIAVERDXIVYERDIAN
130
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<iomanip>
#include<list>
#include<deque>
#include<map>
#include <stdio.h>
#include <queue>
#define maxn 10000+5
#define ull unsigned long long
#define ll long long
#define reP(i,n) for(i=1;i<=n;i++)
#define rep(i,n) for(i=0;i<n;i++)
#define cle(a) memset(a,0,sizeof(a))
#define mod 90001
#define PI 3.141592657
#define INF 1<<30
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;
bool cmp(int a,int b){
return a>b;
}
char t[1000005];
char p[10005];
int f[10005];
void getFail(char *p,int *f)
{
int i,j;
f[0]=0;
f[1]=0;
int m=strlen(p);
for(int i=1;i<m;i++)
{
j=f[i];
while(j&&p[i]!=p[j])
{
j=f[j];
}
f[i+1]=p[i]==p[j]?j+1:0;
}
}
int find(char * t,char *p,int *f)
{
int sum=0;
int n=strlen(t),m=strlen(p);
getFail(p,f);
int j=0;
for(int i=0;i<n;i++)
{
while(j&&p[j]!=t[i])
{
j=f[j];
}
if(p[j]==t[i])
j++;
if(j==m)
{
sum++;
j=f[j];///回溯找到上一个失配点(对称性)
}
}
return sum;
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
int n;
cin>>n;
while(n--)
{
scanf("%s",&p);
scanf("%s",&t);
cout<<find(t,p,f)<<endl;
}
return 0;
}
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