codeforces 831A Unimodal Array

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最新推荐文章于 2021-10-19 19:39:57 发布

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本文介绍了一个简单的算法来判断一个整数数组是否符合单峰数组的特点：即数组元素开始严格递增，随后可能保持不变，最后严格递减。通过示例说明了单峰数组的定义，并给出了实现这一功能的C++代码。

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Array of integers is unimodal, if:

it is strictly increasing in the beginning;
after that it is constant;
after that it is strictly decreasing.

The first block (increasing) and the last block (decreasing) may be absent. It is allowed that both of this blocks are absent.

For example, the following three arrays are unimodal: [5, 7, 11, 11, 2, 1], [4, 4, 2], [7], but the following three are not unimodal: [5, 5, 6, 6, 1], [1, 2, 1, 2], [4, 5, 5, 6].

Write a program that checks if an array is unimodal.

Input

The first line contains integer n (1 ≤ n ≤ 100) — the number of elements in the array.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1 000) — the elements of the array.

Output

Print "YES" if the given array is unimodal. Otherwise, print "NO".

You can output each letter in any case (upper or lower).

        Examples
      
          input
        
6
1 5 5 5 4 2

          output
        
YES

          input
        
5
10 20 30 20 10

          output
        
YES

          input
        
4
1 2 1 2

          output
        
NO

          input
        
7
3 3 3 3 3 3 3

          output
        
YES

Note

In the first example the array is unimodal, because it is strictly increasing in the beginning (from position 1 to position 2, inclusively), that it is constant (from position 2 to position 4, inclusively) and then it is strictly decreasing (from position 4 to position 6, inclusively).

总是写的不对，后来看了一下别人的，就是判断一个数组是不是只有一个峰

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
    int n,a[110];
    scanf("%d",&n);
    memset(a,1010,sizeof(a));
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    int t=1;
    while(t<=n-1&&a[t]<a[t+1])
        t++;
    while(t<=n-1&&a[t]==a[t+1])
        t++;
    for(int i=t;i<=n-1;i++)
    {
        if(a[i]<=a[i+1])
        {
            printf("NO\n");
            return 0;
        }
    }
    printf("YES\n");
    return 0;
}