1、http://acm.hdu.edu.cn/showproblem.php?pid=1686
2、题目大意：
这道题目太纠结了，就是一道直接用KMP模板的题目，比赛的时候居然第一次想成做for循环次的KMP，超时了，后来知道只能用一次KMP了，结果用的string 一直超时，后来改成char居然一遍AC，还得好好熟悉算法
题目是给出两个字符串a,b,求a在b中出现多少次
3、题目：
Oulipo
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3999 Accepted Submission(s): 1578
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.
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.
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
Source
华东区大学生程序设计邀请赛_热身赛
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4、ＡＣ代码：
4、ＡＣ代码：
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<algorithm>
using namespace std;
char s[1000005];
char a[10005];
char b[1000005];
int f[10005];
void getFail(char p[], int *f)
{int 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]?1+j:0;}
}
int find(char T[],char p[],int *f)
{getFail(p,f);int n=strlen(T);int m=strlen(p);int j=0;int sum=0;for(int i=0; i<n; ++i){while(j && T[i]!=p[j])j=f[j];if(T[i]==p[j])++j;if(j==m){sum++;}}return sum;
}
int main()
{int n;scanf("%d",&n);while(n--){scanf("%s",a);scanf("%s",b);int la=strlen(a);int lb=strlen(b);int sum=find(b,a,f);printf("%d\n",sum);}return 0;
}
附string 超时代码：
#include<iostream>
#include<string.h>
#include<string>
#include<algorithm>
using namespace std;
char s[1000005];
string a;
string b;
int f[10005];
void getFail(string p, int *f)
{int m=p.length();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]?1+j:0;}
}
int find(string T,string p,int *f)
{getFail(p,f);int n=T.length();int m=p.length();int j=0;int sum=0;for(int i=0; i<n; ++i){while(j && T[i]!=p[j])j=f[j];if(T[i]==p[j])++j;if(j==m){sum++;}}return sum;
}
int main()
{int n;cin>>n;while(n--){cin>>a;cin>>b;int la=a.length();int lb=b.length();int sum=find(b,a,f);//printf("%d\n",sum);cout<<sum<<endl;}return 0;
}