CodeForces - 831A Unimodal Array

最新推荐文章于 2019-02-04 10:11:44 发布

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最新推荐文章于 2019-02-04 10:11:44 发布

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本文介绍了一种算法，用于判断一个整数数组是否符合有趣数组的定义：即数组开始部分严格递增，随后可能有一段常量部分，最后严格递减。文中提供了一个示例代码，展示了如何通过三次遍历来检查数组是否满足这些条件。

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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
inputCopy
6
1 5 5 5 4 2
outputCopy
YES
inputCopy
5
10 20 30 20 10
outputCopy
YES
inputCopy
4
1 2 1 2
outputCopy
NO
inputCopy
7
3 3 3 3 3 3 3
outputCopy
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).

/*如果数组满足以下三个条件，则这个数组是有趣的：

它刚开始是严格递增的
之后是不变的
再之后它是严格递减的

但除了第二个条件，其他两个条件（递增和递减）不存在的数组也可以是有趣的。

例如，以下三个数组是有趣的： [5, 7, 11, 11, 2, 1], [4, 4, 2], [7], 但以下三个却不是有趣的: [5, 5, 6, 6, 1], [1, 2, 1, 2], [4, 5, 5, 6].

请您编写一个程序判断一个数组是否有趣*/

直接模拟能到达最后就是 YES 否则 NO

#include <iostream>
using namespace std;
int main()
{
	int n,a[1005];
	cin>>n;
	for(int i=0;i<n;i++)
	{
		cin>>a[i];
	 } 
	 int i=1;
	 for(;i<n;) 
	 {
	 	if(a[i]>a[i-1]) i++;
	 	else break;
	 }
	 for(;i<n;)
	 {
	 	if(a[i]==a[i-1]) i++;
	 	else break;
	 }
	 for(;i<n;)
	 {
	 	if(a[i]<a[i-1]) i++;
	 	else break;
	 }
	 if(i!=n) cout<<"NO"<<endl;
	 else cout<<"YES"<<endl;
	 return 0;
 }