CodeForces 831A：Unimodal Array（模拟）

Unimodal Array

Time limit:1000 ms Memory limit:262144 kB OS:Linux

---

Problem Description

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).

Example

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).

---

题意：

给出一个序列，判断其是否是按照：严格递增，相等，严格递减的格式，其中第一段和第三段可以没有

解题思路：

找到最大的那个数字的下标，然后往左判断是否严格递减，然后走到最后一个相同的元素，往右判断是否严格递减

---

Code：

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#define mem(a,b) memset(a,b,sizeof(a))using namespace std;const int maxn=100+5;int a[maxn];int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        int m=0;        for(int i=0; i<n; i++)        {            scanf("%d",a+i);            if(a[m]<a[i])                m=i;        }        int flag=1;        for(int i=m-1; i>=0; i--)        {            if(a[i]>=a[i+1])            {                flag=0;                break;            }        }        while(a[m]==a[m+1]&&m<n-1)            m++;        for(int i=m+1; i<n; i++)        {            if(a[i]>=a[i-1])            {                flag=0;                break;            }        }        if(flag)            printf("YES\n");        else            printf("No\n");    }    return 0;}