Oulipo 、Power Strings (KMP入门）

题目：

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

3BAPCBAPCAZAAZAZAZAVERDIAVERDXIVYERDIAN

Sample Output

130

参考代码：

#include<cstdio>#include<cstring>using namespace std;int next[1000000];char w[10000],t[1000000];void getNext(){int i = 0,j = -1;next[0] = -1;while(w[i]){if(j == -1 || w[i] == w[j]){++ i,++j;if(w[i] == w[j])next[i] = next[j];else next[i] = j;}elsej = next[j];}}int find(){int ans = 0;getNext();int i = 0,j = 0;while(t[i]){if(j == -1 || t[i] == w[j])++ i,++ j;elsej = next[j];if(j != -1 && !w[j]){ans ++;j = next[j];}}return ans;}int main(){int cases;scanf("%d",&cases);while(cases --){scanf("%s",w);scanf("%s",t);printf("%d\n",find());}}

http://blog.csdn.net/kakaxi2222/article/details/73826590

next数组当不匹配的时候返回的位置是:

Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).

Input

Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.

Output

For each s you should print the largest n such that s = a^n for some string a.

Sample Input

abcdaaaaababab.

Sample Output

143

参考代码：

#include<cstdio>#include<cstring>char s[100000000];int next[100000000];int main(){while(scanf("%s",s),s[0]!='.'){int i,j,len = strlen(s);next[0] = next[1] = 0;for(i = 1;i < len;i ++){j = next[i];while(j && s[i]!= s[j])j = next[j];next[i+1] = s[j]==s[i]?j+1:0;}if(len % (len-next[len]) == 0)printf("%d\n",len /(len - next[len]));elseprintf("1\n");}return 0;}