题意：字符串匹配：寻找字符串S中，字符串T出现的次数

思路：KMP或哈希

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

代码：KMP

/*next数组的含义就是一个固定字符串的最长前缀和最长后缀相同的长度。
ABCEFGHRHHABCABCEFGHRHHABC
末尾（ABC）能跟开头（ABC）重合的最大长度就是前缀和子缀的最大公共长度*/
#include<stdio.h>
#include<string.h>
using namespace std;
int n,a,b,dp[10010];
char u[10010],v[1000010];
void ne()//**注意最长前缀：是说以第一个字符开始，但是不包含最后一个字符。*/
{int i=0,j=-1;dp[0]=-1;//next[0]初始化为-1，-1表示不存在相同的最大前缀和最大后缀while(i<a){if(j==-1||u[i]==u[j]) dp[++i]=++j;//这个是把算的k的值（就是相同的最大前缀和最大后缀长）赋给next[q]else j=dp[j];//如果下一个不同，那么k就变成next[k]，注意next[k]是小于k的，无论k取任何值。}//往前回溯
}
void KMP()
{int ans=0;int i=0,j=0;while(i<a&&j<b){if(i==-1||u[i]==v[j])j++,i++;else i=dp[i];if(i==a){ans++;i=dp[i];}}printf("%d\n",ans);
}
int main()
{scanf("%d",&n);while(n--){scanf("%s%s",u,v);a=strlen(u);b=strlen(v);ne();KMP();}return 0;
}

哈希

#include<stdio.h>/*哈希*/
#include<string.h>
#include<algorithm>
using namespace std;
typedef unsigned long long ull;
const int M=1e4+10;
const int p=13331;/*哈希的基数*/
int t,a,b;
char u[M],v[100*M];
ull dp[M];
ull h[M*100];
ull k;
void init()
{dp[0]=1;for(int i=1;i<M;i++)dp[i]=dp[i-1]*p;
}
ull geth(char *s)
{ull h=0;int len=strlen(s);for(int i=0;i<len;i++)h=h*p+s[i];return h;
}
void gethash(char *s)
{int len=strlen(s);h[0]=0;for(int i=0;i<len;i++)h[i+1]=h[i]*p+s[i];
}int main()
{scanf("%d",&t);init();while(t--){scanf("%s%s",u+1,v+1);k=geth(u+1);gethash(v+1);a=strlen(u+1);b=strlen(v+1);int ans=0;for(int i=0;i+a<=b;i++){ull tmp=h[i+a]-h[i]*dp[a];if(tmp==k)ans++;}printf("%d\n",ans);}return 0;
}