B - Vladik and Complicated Book（思维题）-白红宇的个人博客

发布日期：2021-10-16 05:05:03 浏览次数：12 分类：技术文章

本文共 2697 字，大约阅读时间需要 8 分钟。

Vladik had started reading a complicated book about algorithms containing n pages. To improve understanding of what is written, his friends advised him to read pages in some order given by permutation P = [p₁, p₂, ..., p_n], where p_i denotes the number of page that should be read i-th in turn.

Sometimes Vladik’s mom sorted some subsegment of permutation P from position l to position r inclusive, because she loves the order. For every of such sorting Vladik knows number x — what index of page in permutation he should read. He is wondered if the page, which he will read after sorting, has changed. In other words, has p_x changed? After every sorting Vladik return permutation to initial state, so you can assume that each sorting is independent from each other.

Input

First line contains two space-separated integers n, m (1 ≤ n, m ≤ 10⁴) — length of permutation and number of times Vladik's mom sorted some subsegment of the book.

Second line contains n space-separated integers p₁, p₂, ..., p_n (1 ≤ p_i ≤ n) — permutation P. Note that elements in permutation are distinct.

Each of the next m lines contains three space-separated integers l_i, r_i, x_i (1 ≤ l_i ≤ x_i ≤ r_i ≤ n) — left and right borders of sorted subsegment in i-th sorting and position that is interesting to Vladik.

Output

For each mom’s sorting on it’s own line print "Yes", if page which is interesting to Vladik hasn't changed, or "No" otherwise.

Example

Input

5 55 4 3 2 11 5 31 3 12 4 34 4 42 5 3

Output

YesNoYesYesNo

Input

6 51 4 3 2 5 62 4 31 6 24 5 41 3 32 6 3

Output

YesNoYesNoYes

Note

Explanation of first test case:

[1, 2, 3, 4, 5] — permutation after sorting, 3-rd element hasn’t changed, so answer is "Yes".

[3, 4, 5, 2, 1] — permutation after sorting, 1-st element has changed, so answer is "No".

[5, 2, 3, 4, 1] — permutation after sorting, 3-rd element hasn’t changed, so answer is "Yes".

[5, 4, 3, 2, 1] — permutation after sorting, 4-th element hasn’t changed, so answer is "Yes".

[5, 1, 2, 3, 4] — permutation after sorting, 3-rd element has changed, so answer is "No".

题意：

有一串数字，然后给你 l r分别代表左边位子和右边位子，将他们中间的数排序即可。然后判断他们中间某一位数字是否发生了变化。

思路：

一开始没太读懂题意，只是看了数据以为是将他们之间的数翻转，其实不是，而是由小到大进行排序，所以wa。。。后来发现原因，用的sort快排，但是遗憾的是 tle...

然后又重新整理了思路，发现只要在这个区间的小与等于x这位数的个数和等于x到l之间的个数和加1，因为x这个数本身也在所以加1.即s=x-l+1；

代码：

#include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        using namespace std;int a[20000],temp[20000];int main(){    ios::sync_with_stdio(false);  int n,m,i;  int l,r,x;  int s=0;  while(cin>>n>>m)  {      for(i=1;i<=n;i++)cin>>a[i];      while(m--)      {          s=0;          //for(i=0;i
        
         >l>>r>>x; for(i=l;i<=r;i++)if(a[x]>=a[i])s++; if(s==x-l+1)cout<<"Yes"<

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