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题目描述

给定一个含n个数的序列A和一个含m (m<=n) 个数的序列B。

询问在A中有多少段连续的长为m的子序列Ak,Ak+1,…,Ak+m-1使得对于任意1<=i, j<=m满足Ak+i-1-Bi=Ak+j-1-Bj

输入

第一行两个整数n，m (1 <= m <= n <= 106)

接下来一行n个整数，描述序列A (Ai <= 109)

接下来一行m个整数，描述序列B (Bi <= 109)

输出

输出一个数表示答案

样例输入

7 4
6 6 8 5 5 7 4
7 7 9 6

样例输出

2

思路

KMP模板题

AC

#include<iostream>
#include<bits/stdc++.h>
#define N 1000006
using namespace std;
int a[N], b[N], Next[N];
void get_next(int m) {  //求Next数组 
    Next[0] = -1;    
    int i = 0, j = -1;
    while (i < m) {
        if(j == -1 || b[i] == b[j]) {
            Next[++i] = ++j;    //赋值 
        }else {
            j = Next[j];    //回溯 
        }
    }
}
int kmp(int n, int m) {
    get_next(m);    //求Next数组 
    int i = 0, j = 0;
    int ans = 0;
    while (i < n) {
        if (j == -1 || a[i] == b[j]) {  //当前匹配成功进行下一个匹配 
            i++;
            j++;    
        }else {
            j = Next[j];
        }
        if (j == m) {   //匹配成功 
            ans++;
            j = Next[j];    //进行下一次匹配 
        }
    }
    return ans;
}

int main() {
// freopen("in.txt", "r", stdin);
    int n, m;
    while (scanf("%d%d", &n, &m) != EOF) {
        for (int i = 0; i < n; i++) {
            scanf("%d",&a[i]);
        }
        for (int i = 0; i < m; i++) {
            scanf("%d", &b[i]);
        }
        if (m == 1) {   //m = 1 特判 
            printf("%d\n", n);
            continue;
        }
        for (int i = 1; i < n; i++) {
            a[i - 1] = a[i] - a[i - 1];
        }
        for (int i = 1; i < m; i++) {
            b[i - 1] = b[i] - b[i - 1];
        }
        printf("%d\n", kmp(n - 1, m - 1));      
    }

    return 0;
}