1. 题目

1.1 题目链接

http://poj.org/problem?id=3461
类似题目：LeetCode 30. 串联所有单词的子串（字符串哈希）

1.2 题目大意

模式串在主串中出现过几次。

2. Accepted代码

2.1 KMP解法

#include <stdio.h>
#include <iostream>
#include <cstring>
int next[10001];
char a[1000001], b[10001];
void calNexts(char *b, int m, int *next)
{next[0] = -1;int j = 0, k = -1;while(j < m){if(k == -1 || b[j] == b[k]){j++;k++;next[j] = k;}elsek = next[k];}
}
int str_kmp(char *a, int n, char *b, int m)
{calNexts(b, m, next);int i = 0, j = 0, times = 0;while(i < n && j < m){if(j == -1 || a[i] == b[j]){i++;j++;}else{j = next[j];}if(j == m){times++;j = next[j];}}return times;
}
int main()
{int count;std::cin >> count;while(count--){scanf("%s",&b);scanf("%s",&a);printf("%d\n",str_kmp(&a[0], strlen(a), &b[0], strlen(b)));}return 0;
}

2.2 哈希法（有推导过程）

参考别人的哈希函数，哈希值的求法很犀利
自己推导的过程如下：

/*** @description: poj 3461 字符串匹配，哈希法* @author: michael ming* @date: 2019/6/26 21:59* @modified by: */
#include <stdio.h>
#include <iostream>
#include <cstring>
#include <cmath>
#define K 25    //K大于等于字符集数 k>=25
typedef unsigned long long ull;
char a[1000001], b[10001];
ull ha[1000001];
ull powk[10001];//存储K的幂值
using namespace std;
void fillpowk(int m)
{powk[0] = 1;for(int i = 1; i <= m; ++i)powk[i] = powk[i-1]*K;
}
ull hashf(int m, char *str)
{ull value = 0;for(int i = 0; i < m; ++i)value = value * K + str[i];return value;
}
int str_RK(char *a, int n, char *b, int m)//a是主串，b是模式串
{fillpowk(m);int i, times = 0;ull hash_val, value = 0;value = hashf(m,b); //计算模式串的hash值valueha[0] = ull(a[0]);for(i = 1; i < n; ++i)ha[i] = ha[i-1] * K + a[i];for(i = 0; i < n-m+1; ++i)//最多n-m+1次比较{if(i == 0)hash_val = hashf(m,a);elsehash_val = ha[i+m-1] - ha[i-1]*powk[m];//这个公式可以自己推导一下if(hash_val == value){//如果子串哈希值等于模式串的，匹配成功（该哈希无冲突）times++;}}return times;
}
int main()
{int count;std::cin >> count;while(count--){scanf("%s",&b);scanf("%s",&a);printf("%d\n",str_RK(&a[0], strlen(a), &b[0], strlen(b)));}return 0;
}