CF 171 （div2）C

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 integers
a1, a2, ..., an. Also, you've got
m queries, the i-th query is described by two integers
li, ri. Numbers
li, ri 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, ..., bk, 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 ≤ ... ≤ bx ≥ bx + 1 ≥ bx + 2... ≥ bk.
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, ..., an
(1 ≤ ai ≤ 109), where number
ai 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 li,
ri
(1 ≤ li ≤ ri ≤ 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.

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

其实是个简单题，思维混乱时容易想不清楚。简单的预处理出每个点往左和往右的最长不递减序列末尾的编号就可以了。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <cmath>
#define LL long long
using namespace std;
const int maxn = 100000 + 5;

int dp[maxn];
int num[maxn],r[maxn],l[maxn];

int main(){
    int n,m,last;
    while(cin >> n >> m){
        for(int i = 1;i <= n;i++) cin >> num[i];
        last = 1;
        for(int i = 2;i <= n;i++){
            if(num[i] > num[i-1]){
                for(int j = last;j < i;j++) l[j] = last;
                last = i;
            }
            if(i == n){
                for(int j = last;j <= n;j++) l[j] = last;
            }
        }
        last = n;
        for(int i = n-1;i >= 1;i--){
            if(num[i] > num[i+1]){
                for(int j = last;j > i;j--) r[j] = last;
                last = i;
            }
            if(i == 1){
                for(int j = last;j >= 1;j--) r[j] = last;
            }
        }
        for(int i = 0;i < m;i++){
            int x,y;
            cin >> x >> y;
            if(r[x] >= l[y]) cout << "Yes\n";
            else cout << "No\n";
        }
    }
    return 0;
}