CodeForces 471D MUH and Cube Walls

本文介绍了一个CodeForces平台上的题目,该题目通过KMP算法解决两个塔堆匹配的问题,具体为在一排不同高度的塔堆中寻找与特定序列相匹配的连续塔堆段。

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题目链接:http://codeforces.com/problemset/problem/471/D


MUH and Cube Walls

time limit per test :2 seconds
memory limit per test :256 megabytes
input :standard input
output :standard output

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".

Input

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 ai (1 ≤ ai ≤ 109) — the heights of the towers in the bears' wall. The third line contains w integers bi (1 ≤ bi ≤ 109) — the heights of the towers in the elephant's wall.

Output

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

Examples

Input
13 5
2 4 5 5 4 3 2 2 2 3 3 2 1
3 4 4 3 2
Output
2

Note

The picture to the left shows Horace's wall from the sample, the picture to the right shows the bears' wall. The segments where Horace can "see an elephant" are in gray.


思路:其实就是KMP算法的模板题。关键在于要能够想到相邻两个数作差处理。只要算出全部两个相邻数之间的差,那么剩下的就是裸的KMP算法求匹配串的个数问题了。为什么要作差而不是别的呢?请看下面。

假设第一个数组是a,b,c,第二个数组是d,e,f,那么满足题意就必须满足d-a=e-b=f-c,即e-d=b-a且c-b=f-e,所以要作差处理。注意:m=1时要特殊处理一下。


附上AC代码:

#include <bits/stdc++.h>
using namespace std;
const int maxn = 200005;
int t[maxn], p[maxn], f[maxn];
int n, m;

void get_fail(){
	f[0] = f[1] = 0;
	for (int i=1; i<m; ++i){
		int j = f[i];
		while (j && p[i]!=p[j])
			j = f[j];
		f[i+1] = p[i]==p[j] ? j+1 : 0;
	}
}

int kmp(){
	get_fail();
	int j=0, cnt=0;
	for (int i=0; i<n; ++i){
		while (j && p[j]!=t[i])
			j = f[j];
		if (p[j] == t[i])
			++j;
		if (j == m)
			++cnt;
	}
	return cnt;
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin >> n >> m;
	for (int i=0; i<n; ++i)
		cin >> t[i];
	for (int i=0; i<m; ++i)
		cin >> p[i];
	if (m == 1){
		cout << n << endl;
		return 0;
	}
	for (int i=0; i<n-1; ++i)
		t[i] = t[i+1]-t[i];
	for (int i=0; i<m-1; ++i)
		p[i] = p[i+1]-p[i];
	--n, --m;
	cout << kmp() << endl;
	return 0;
}


### Codeforces 1487D Problem Solution The problem described involves determining the maximum amount of a product that can be created from given quantities of ingredients under an idealized production process. For this specific case on Codeforces with problem number 1487D, while direct details about this exact question are not provided here, similar problems often involve resource allocation or limiting reagent type calculations. For instance, when faced with such constraints-based questions where multiple resources contribute to producing one unit of output but at different ratios, finding the bottleneck becomes crucial. In another context related to crafting items using various materials, it was determined that the formula `min(a[0],a[1],a[2]/2,a[3]/7,a[4]/4)` could represent how these limits interact[^1]. However, applying this directly without knowing specifics like what each array element represents in relation to the actual requirements for creating "philosophical stones" as mentioned would require adjustments based upon the precise conditions outlined within 1487D itself. To solve or discuss solutions effectively regarding Codeforces' challenge numbered 1487D: - Carefully read through all aspects presented by the contest organizers. - Identify which ingredient or component acts as the primary constraint towards achieving full capacity utilization. - Implement logic reflecting those relationships accurately; typically involving loops, conditionals, and possibly dynamic programming depending on complexity level required beyond simple minimum value determination across adjusted inputs. ```cpp #include <iostream> #include <vector> using namespace std; int main() { int n; cin >> n; vector<long long> a(n); for(int i=0;i<n;++i){ cin>>a[i]; } // Assuming indices correspond appropriately per problem statement's ratio requirement cout << min({a[0], a[1], a[2]/2LL, a[3]/7LL, a[4]/4LL}) << endl; } ``` --related questions-- 1. How does identifying bottlenecks help optimize algorithms solving constrained optimization problems? 2. What strategies should contestants adopt when translating mathematical formulas into code during competitive coding events? 3. Can you explain why understanding input-output relations is critical before implementing any algorithmic approach? 4. In what ways do prefix-suffix-middle frameworks enhance model training efficiency outside of just tokenization improvements? 5. Why might adjusting sample proportions specifically benefit models designed for tasks requiring both strong linguistic comprehension alongside logical reasoning skills?
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