2018.2.4【 CodeForces - 831A 】解题报告（模拟）

A. Unimodal Array

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

【题目大意】

一个数列按照递增->不变->递减的先后顺序，是符合要求的数列。可以缺少某一部分，比如先增后减没有相等部分，或者只有相等和减少。

符合要求的输出YES，不符合的输出NO。

【解题思路】

模拟，我的方法是用一个数组来储存前一个数到下一个数的增减情况。a[i+1]>a[i]，则b[i]=1,a[i+1]==a[i]，则b[i]=2，a[i+1]

【解题代码】

#include 
#include 
#include 
#include 
#include 
#define maxn 10010
using namespace std;
int array[maxn];
int used[maxn];
int main()
{
	int n;
	while(~scanf("%d",&n))
	{
	
	for(int i=0;iarray[i])//increase
		{
			used[i]=1;
		} 
		else if(array[i+1]==array[i]) //const
		{
			used[i]=2;
		}
		else//decrease
		{
			used[i]=3;
		} 
	}
	int i=0;
	for(;i+1used[i+1])
			break;
	}
	if(n==1) printf("YES\n");
	else if(i!=n-2) printf("NO\n");
	else printf("YES\n");
	}
	return 0;
}

【收获与反思】

用另一个数组储存变化状态。