[codeforces471D]MUH and Cube Walls

[codeforces471D]MUH and Cube Walls

试题描述

Polar bears Menshykov and Uslada from the zoo of St. Petersburg and elephant Horace from the zoo of Kiev got hold of lots of wooden cubes somewhere. They started making cube towers by placing the cubes one on top of the other. They defined multiple towers standing in a line as a wall. A wall can consist of towers of different heights.

Horace was the first to finish making his wall. He called his wall an elephant. The wall consists of w towers. The bears also finished making their wall but they didn't give it a name. Their wall consists of n towers. Horace looked at the bears' tower and wondered: in how many parts of the wall can he "see an elephant"? He can "see an elephant" on a segment of w contiguous towers if the heights of the towers on the segment match as a sequence the heights of the towers in Horace's wall. In order to see as many elephants as possible, Horace can raise and lower his wall. He even can lower the wall below the ground level (see the pictures to the samples for clarification).

Your task is to count the number of segments where Horace can "see an elephant".

输入

The first line contains two integers n and w (1 ≤ n, w ≤ 2·10⁵) — the number of towers in the bears' and the elephant's walls correspondingly. The second line contains n integers a_i (1 ≤ a_i ≤ 10⁹) — the heights of the towers in the bears' wall. The third line contains w integers b_i (1 ≤ b_i ≤ 10⁹) — the heights of the towers in the elephant's wall.

输出

Print the number of segments in the bears' wall where Horace can "see an elephant".

输入示例

13 5
2 4 5 5 4 3 2 2 2 3 3 2 1
3 4 4 3 2

输出示例

2

数据规模及约定

见“输入”

题解

两个序列差分一下后跑 KMP。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
using namespace std;

int read() {
	int x = 0, f = 1; char c = getchar();
	while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
	while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
	return x * f;
}

#define maxn 200010
int n, m, S[maxn], T[maxn], Fail[maxn];

int main() {
	n = read(); m = read();
	for(int i = 1; i <= n; i++) S[i] = read();
	for(int i = 1; i <= m; i++) T[i] = read();
	
	if(m == 1) return printf("%d
", n), 0;
	
	for(int i = 1; i < n; i++) S[i] = S[i+1] - S[i]; n--;
	for(int i = 1; i < m; i++) T[i] = T[i+1] - T[i]; m--;
	for(int i = 2; i <= m + 1; i++) {
		int j = Fail[i-1];
		while(j > 1 && T[j] != T[i-1]) j = Fail[j];
		Fail[i] = T[j] == T[i-1] ? j + 1 : 1;
	}
	int p = 1, ans = 0;
	for(int i = 1; i <= n; i++) {
		while(p > 1 && T[p] != S[i]) p = Fail[p];
		if(T[p] == S[i] && p == m) ans++;
		p = T[p] == S[i] ? p + 1 : 1;
	}
	
	printf("%d
", ans);
	
	return 0;
}