（CodeForces - 831A）Unimodal Array

最新推荐文章于 2019-02-04 10:11:44 发布

原创最新推荐文章于 2019-02-04 10:11:44 发布 · 360 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#codeforces

模拟专栏收录该内容

15 篇文章

订阅专栏

本文介绍了一种用于判断数字序列是否为unimodal（即序列先递增、后不变或递减）的算法实现，并提供了完整的C++代码示例。通过对输入序列的增减特性进行模拟检查，该算法能够有效地确定序列是否符合unimodal定义。

（CodeForces - 831A）Unimodal Array

time limit per test: 1 second

memory limit per test: 256 megabytes

inputs: tandard 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

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).

题目大意：一个数字序列若满足：开始递增，中间保持不变，后面递减，就说这个序列是unimodal，前面的递增和后面递减可以省掉，给出一个序列问他是不是unimodal。

思路：模拟序列的增减性即可。

#include<cstdio>
using namespace std;

const int INF=0x3f3f3f3f;
const int maxn=105;
int a[maxn];

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++) scanf("%d",a+i);
        a[n]=INF;
        int tot=0;
        while(a[tot+1]>a[tot]) tot++;
        while(a[tot+1]==a[tot]) tot++;
        while(a[tot+1]<a[tot]) tot++;
        if(tot<n-1) printf("NO\n");
        else printf("YES\n");   
    }
    return 0;
}