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
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<stack>
#include<queue>
#include<bitset>
#include<set>
#include<map>
#define inf 0x3f3f3f3f
#define ll long long
#define pi acos(-1)
using namespace std;
char mo[1000009];
int next1[1000009];
char str[1000009];
void getnext()
{
next1[0] = -1;
int i = 0,j = -1;
int len = strlen(mo);
while(i < len)
{
if(j == -1 || mo[i] == mo[j])//如果是第一次或要進行的和他前面的指定位置相同
{
i++;
j++;//主串和副串一起前進
next1[i] = j;//記錄匹配長度
}
else{
j = next1[j];//不行找他記錄的上一個可以匹配到的
}
}
}//對模式串進行操作
int kmp()
{
int ans = 0;
int i = 0,j = 0;
int len1 = strlen(mo);
int len2 = strlen(str);
while(i < len2)
{
if(mo[j] == str[i] || j == -1)
{
i++;j++;
}
else{
j = next1[j];
}
if(j == len1)
{//這裏是找到匹配位置的地方，i存在主串的位置用時搞就行
ans++;//記錄有幾次匹配
j = next1[j]; //繼續執行
}
}
return ans;//返回幾次
}
int main()
{
int n;
scanf("%d",&n);
while(n--)
{
scanf("%s",&mo);//存好模式串
scanf("%s",&str);//主串（最長的 ）
getnext();//搞出next數組
printf("%d\n",kmp());//kmp輸出主串中有幾個匹配串
}
return 0;
}