CodeForce 279C

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：

C. Ladder

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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_li, a_li + 1, a_li + 2, ..., a_ri. 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 thei-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 61 2 1 3 3 5 2 11 32 32 48 81 45 8

Output

YesYesNoYesNoYes

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");elseprintf("No\n");}return 0;}