本文探讨了法国作家Georges Perec如何在作品《Ladisparition》中仅使用非'e'字母，进而引出了一个关于文字游戏与算法的有趣挑战。通过设置限制条件，要求参赛者用尽可能少的指定词汇来创作有意义的文本，本文将介绍一种算法——KMP算法，用于高效地计算特定词汇在长文本中的出现次数。该算法特别适用于处理大量文本数据，如参赛作品，从而为评选提供准确的排名依据。
Oulipo
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1677 Accepted Submission(s): 646
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
3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN
Sample Output
1
3
0
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1677 Accepted Submission(s): 646
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
3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN
Sample Output
1
3
0
题意就是找模式串在母串中出现的次数，，，kmp算法。。。这里是从1开始，比较方便。。。
代码：
#include<iostream> #include<cstdio> #include<string.h> #define N 10005 #define M 1000005 using namespace std; char a[N],s[M]; int next[N]; int n,m; void KMP_next() { a[0]='#';//这里是从a+1开始输入，一定要注意，给a[0]赋值 n=strlen(a); m=strlen(s); next[1]=0; int j=0; for(int i=2;i<n;i++) { while(j>0&&a[j+1]!=a[i]) j=next[j]; if(a[j+1]==a[i]) j++; next[i]=j; } } int KMP_match() { int sum=0,i=0,j=0; for(int i=0;i<m;++i) { while(j>0&&a[j+1]!=s[i]) j=next[j]; if(a[j+1]==s[i]) j++; if(j==n-1) { sum++; j=next[j]; } } return sum; } int main() { int Case; scanf("%d",&Case); while(Case--) { //getchar(); scanf("%s%s",a+1,s); KMP_next(); printf("%d\n",KMP_match()); } return 0; }这个超时的不知为什么，大牛看到时指点一下不胜感激，，，
代码：
#include<iostream> #include<string> #define N 10005 using namespace std; int next[N]; int KMP_next(string &a,string &b) { int n=a.size(); int m=b.size(); int j=0,i=1,sum=0; next[0]=0; while(i<n) { if(a[j]==a[i]) { next[i]=j+1; j++;i++; } else{ if(j>0) j=next[j-1]; else next[i++]=0; } } i=0,j=0; while(i<m) { if(a[j]==b[i]) { if(j==n-1) {sum++;j=next[j-1]; } else {j++;i++;} } else { if(j>0) j=next[j-1]; else i++; } } return sum; } int main() { int Case; cin>>Case; while(Case--) { string a,b; cin>>a>>b; cout<<KMP_next(a,b)<<endl; } return 0; }
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