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.

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.

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.

////  main.cpp//  poj3461////  Created by Candy on 10/19/16.//  Copyright © 2016 Candy. All rights reserved.//#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N=1e6+5,M=1e4+5;int T,f[M];// already i(0...i-1),now pos ichar s[N],p[M];void getFail(){    size_t m=strlen(p);    f[0]=f[1]=0;    for(int i=1;i<m;i++){        int j=f[i];        while(j&&p[i]!=p[j]) j=f[j];        f[i+1]=p[i]==p[j]?j+1:0;    }}int kmp(){    getFail();    int cnt=0;    size_t n=strlen(s),m=strlen(p);    int j=0;    for(int i=0;i<n;i++){        while(j&&s[i]!=p[j]) j=f[j];        if(s[i]==p[j]) j++;        if(j==m) cnt++;    }    return cnt;}int main(int argc, const char * argv[]) {    scanf("%d",&T);    while(T--){        scanf("%s%s",p,s);        printf("%d\n",kmp());    }        return 0;}

以前愚蠢的从0开始

从0开始太愚蠢了，于是我从1开始重学重写了一遍

1.算法理解(orz 阮一峰)：http://www.ruanyifeng.com/blog/2013/05/Knuth%E2%80%93Morris%E2%80%93Pratt_algorithm.html

2.摘抄(修改)课件：

• 定义f[i]表示A[1,i]的长度为f[i]的后缀等于A[1,f[i]]  也就是说对于这个子串来说f[i]长度的前缀和后缀相等(注意顺序都是从左到右相等)，那么j+1位置匹配失败就可以转移到f[j]位置(因为1..f[j]和刚刚成功的后缀是一样的)继续匹配f[j]+1
• 多个f[i]符合条件时，取最大的。当然，f[i]是要小于i的，否则就没有意义了。

• 将i从2到n枚举(f[1]=0是不必计算的)，依次计算f[i]。
  计算f[i]时，先令j=f[i-1]，f[i]最大只会是j+1。只需判断A[i]与A[j+1]是否相等就能判断f[i]是不是j+1。若相等，f[i]=j+1；否则令j=f[j]，继续这个过程。
  如果j=0后仍然和j+1不相等，就使f[i]=0

• 时间复杂度O(n)，因为j最多增加减少n次

////  main.cpp//  poj3461////  Created by Candy on 10/19/16.//  Copyright ? 2016 Candy. All rights reserved.//#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N=1e6+5,M=1e4+5;int T,f[M],n,m;char t[N],p[M];void getFail(){    f[1]=0;    for(int i=2;i<=m;i++){        int j=f[i-1];        while(j&&p[i]!=p[j+1]) j=f[j];        f[i]=p[i]==p[j+1]?j+1:0;    }}int kmp(){    int ans=0;    getFail();    int j=0;    for(int i=1;i<=n;i++){        while(j&&t[i]!=p[j+1]) j=f[j];        j+=t[i]==p[j+1];        if(j==m) ans++;    }    return ans;}int main(){//    freopen("in.txt","r",stdin);    scanf("%d",&T);    while(T--){        scanf("%s%s",p+1,t+1);        n=strlen(t+1),m=strlen(p+1);        printf("%d\n",kmp());    }        return 0;}