HDU 1686 Oulipo
2016-02-23 22:06
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本题地址:>here<
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.
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN
3
0
题目形式千变万化,然而它依然是个水题(不就是重复吗)
这是以前的代码,比较乱哈
Problem 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: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.
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
3BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN
Sample Output
13
0
Source
华东区大学生程序设计邀请赛_热身赛题解
我竟无语凝噎……题目形式千变万化,然而它依然是个水题(不就是重复吗)
这是以前的代码,比较乱哈
#include<cstdio> #include<cstring> using namespace std; #define MAX_T 1000010 char st1[MAX_T],st2[MAX_T]; int next1[MAX_T],next2[MAX_T]; int main() { int T; scanf("%d",&T); while(T--) { int ans=0; memset(next1,0,sizeof(next1)); memset(next2,0,sizeof(next2)); scanf("%s%s",&st2[1],&st1[1]); int l1=strlen(&st1[1]),l2=strlen(&st2[1]); int j=0; for(int i=2;i<=l2;i++) { while(j>0&&st2[i]!=st2[j+1]) j=next2[j]; j++; next2[i]=j; if(st2[i]!=st2[j]) next2[i]=0,j=0; } j=0; for(int i=1;i<=l1;i++) { while(j>0&&st1[i]!=st2[j+1]) j=next2[j]; j++; next1[i]=j; if(st1[i]!=st2[j]) next1[i]=0,j=0; } for(int i=1;i<=l1;i++) if(next1[i]==l2) ans++; printf("%d\n",ans); } return 0; }
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