HDU 波峰

Problem @

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 100   Accepted Submission(s) : 41

Problem Description

A sequence of $n$ integers $a_1, a_2, dots, a_n$ is called a peak, if and only if there exists exactly one integer $k$ such that $1 < k < n$, and $a_i < a_{i+1}$ for all $1 le i < k$, and $a_{i-1} > a_i$ for all $k < i le n$.

Given an integer sequence, please tell us if it's a peak or not.

Input

There are multiple test cases. The first line of the input contains an integer $T$, indicating the number of test cases. For each test case:

The first line contains an integer $n$ ($3 le n le 10^5$), indicating the length of the sequence.

The second line contains $n$ integers $a_1, a_2, dots, a_n$ ($1 le a_i le 2 imes 10^9$), indicating the integer sequence.

It's guaranteed that the sum of $n$ in all test cases won't exceed $10^6$.

Output

For each test case output one line. If the given integer sequence is a peak, output "Yes" (without quotes), otherwise output "No" (without quotes).

Sample Input

7
5
1 5 7 3 2
5
1 2 1 2 1
4
1 2 3 4
4
4 3 2 1
3
1 2 1
3
2 1 2
5
1 2 3 1 2

Sample Output

Yes
No
No
No
Yes
No
No

求给定数列是不是波峰，只能是左边递增右边递减

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <cstdio>

using namespace std ; 

#define maxn 1100000
int num[maxn] ; 

int t , n ; 

int main(){
    cin >> t ; 
    while(t--){
        cin >> n ; 
        for(int i=1 ; i<=n ; i++){
            cin >> num[i] ; 
        }
        int sum1 = 0  ; 
        int pos = -1 ; 
        for(int i=2 ; i<n ; i++){
            if(num[i-1] < num[i] && num[i] > num[i+1]){
                sum1 ++ ; 
                pos = i ; 
            }

        }
        bool flag = true ; 
        for(int i=1 ; i<pos ; i++){
            if(num[i] >= num[i+1]){
                flag = false ; 
                break ; 
            }
        }
        for(int i=pos ; i<n ; i++){
            if(flag == false) break ; 
            if(num[i+1]>=num[i]){
                flag = false ;
                break ;  
            }
        }
        if(flag) cout << "Yes" << endl ; 
        else cout << "No" << endl ; 
    }
}