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最新推荐文章于 2022-04-24 22:08:34 发布

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最新推荐文章于 2022-04-24 22:08:34 发布

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分类专栏： ACM_Steps 文章标签： ACM C++

本文链接：https://blog.youkuaiyun.com/unicornet/article/details/8661628

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本文详细介绍了如何使用C语言解决数组序列判断问题，包括输入数组和查询序列，通过计算序列两端递增和递减的元素数量来判断是否构成梯形序列。代码示例清晰易懂。

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好不容易AC了这题(果然是菜鸟-_-|||)，，先记录下来再说。

先来看题目：

You've got an array, consisting of n integersa₁, a₂, ..., a_n. Also, you've gotm queries, the i-th query is described by two integersl_i, r_i. Numbersl_i, r_i define a subsegment of the original array, that is, the sequence of numbersa_{l_i}, a_{l_i + 1}, a_{l_i + 2}, ..., a_{r_i}. For each query you should check whether the corresponding segment is a ladder.

A ladder is a sequence of integers b₁, b₂, ..., b_k, such that it first doesn't decrease, then doesn't increase. In other words, there is such integerx (1 ≤ x ≤ k), that the following inequation fulfills:b₁ ≤ b₂ ≤ ... ≤ b_x ≥ b_x + 1 ≥ b_x + 2... ≥ b_k. Note that the non-decreasing and the non-increasing sequences are also considered ladders.

Input

The first line contains two integers n andm (1 ≤ n, m ≤ 10⁵) — the number of array elements and the number of queries. The second line contains the sequence of integersa₁, a₂, ..., a_n(1 ≤ a_i ≤ 10⁹), where numbera_i stands for thei-th array element.

The following m lines contain the description of the queries. Thei-th line contains the description of the i-th query, consisting of two integers l_i,r_i(1 ≤ l_i ≤ r_i ≤ n) — the boundaries of the subsegment of the initial array.

The numbers in the lines are separated by single spaces.

Output

Print m lines, in the i-th line print word "Yes" (without the quotes), if the subsegment that corresponds to thei-th query is the ladder, or word "No" (without the quotes) otherwise.

Sample test(s)

Input

8 6
1 2 1 3 3 5 2 1
1 3
2 3
2 4
8 8
1 4
5 8

Output

Yes
Yes
No
Yes
No
Yes

大体思路：对用户给出的数组数据逐个分析，记录每个数据ai的两个卫星数据，即在ai左边且递减的数p的个数和在ai右边且递增的数的个数q，然后只要query的长度l - 1小于等于query起始位置的q + 结束位置的p即输出Yes，否则输出No。用一个二维数组来记录p、q，简化计算。

解题代码如下：

#include <stdio.h>
#include <vector>
#include <utility>
using namespace std;
int map[100000][2];
int array[100000];

int main(){
	int n, m;
	scanf("%d%d", &n, &m);
	for (int i = 0; i < n; ++i)
		scanf("%d", &array[i]);
	vector<pair<int, int> > que(m);
	for (int i = 0; i < m; ++i)
		scanf("%d%d", &que[i].first, &que[i].second);
	for (int i = 0; i < n; ++i){
		if (i){
			if (array[i] <= array[i - 1]) map[i][0] = map[i - 1][0] + 1;
			else map[i][0] = 0;
		}
	}
	for (int i = n - 1; i >= 0; --i){
		if (i != n - 1){
			if (array[i] <= array[i + 1]) map[i][1] = map[i + 1][1] + 1;
			else map[i][1] = 0;
		}
	}
	for (int i = 0; i < m; ++i){
		int len = que[i].second - que[i].first;
		if (len <= map[que[i].first - 1][1] || 
			len <= map[que[i].second - 1][0] ||
			len <= map[que[i].first - 1][1] + map[que[i].second - 1][0])
			printf("Yes\n");
		else
			printf("No\n");
	}
	return 0;
}