CF 279C Ladder(区间问题)

• You've got an array, consisting of n integers a1, a2, ..., a**n. Also, you've got m queries, the i-th query is described by two integers l**i, r**i. Numbers l**i, r**i define a subsegment of the original array, that is, the sequence of numbers ali, ali + 1, ali + 2, ..., ari. 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 integer x (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 and m (1 ≤ n, m ≤ 105) — the number of array elements and the number of queries. The second line contains the sequence of integers a1, a2, ..., a**n (1 ≤ a**i ≤ 109), where number a**i stands for the i-th array element.

  The following m lines contain the description of the queries. The i-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 the i-th query is the ladder, or word "No" (without the quotes) otherwise.

Example

• 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
  

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  题意：

  对于一段序列，要求刚开始不能递减，之后不能递增。换言之，只能有一个波峰，不能有波谷，可以是一条平平的线。

  思路：

  算出所求区间左端点向右延伸的距离，右段点向左延伸的距离。两段距离相加，若大于等于区间长度就是yes。

  #include<iostream>
  #include<cstdio> 
  #include<algorithm>
  #include<cmath>
  using namespace std;
  const int mx = 1e5 + 5;
  int num[mx], le[mx], ri[mx];
  int n, m;
  int main(){
  	scanf("%d%d", &n, &m);
  	for(int i = 1; i <= n; i++)
  		scanf("%d", &num[i]); 
  	int a, b ,flag;
  	ri[n] = 0;
  	for(int i = n -1; i >= 1; i--){    //向右延伸的长度 
  		if(num[i] <= num[i + 1])      //等于的时候也是可以延伸的 
  			ri[i] = ri[i + 1] + 1;
  		else ri[i] = 0;
  	}
  	
  	le [1] = 0;
  	for(int i = 2; i <= n; i++){
  		if(num[i] <= num[i -1])     //向左延伸距离 
  			le[i] = le[i -1] + 1;
  		else le[i] = 0; 
  	} 		
  	
  	while(m--){
  		scanf("%d%d", &a, &b);
  		//cout<<ri[a]<<endl;
  	//	cout<<le[b]<<endl;
  		if( (ri[a] + le[b]) >= (b - a))
  			puts("Yes");
  		else 
  			puts("No"); 
  	}
  	
  	return 0;
  }
  

  
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