1 second
256 megabytes
standard input
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.
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.
Print "YES" if the given array is unimodal. Otherwise, print "NO".
You can output each letter in any case (upper or lower).
6
1 5 5 5 4 2
YES
5
10 20 30 20 10
YES
4
1 2 1 2
NO
7
3 3 3 3 3 3 3
YES
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).
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题目的意思是判读啊一个序列是不是先增再平在减的
思路：分类判断
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset> using namespace std; #define LL long long
const int INF = 0x3f3f3f3f;
#define MAXN 2000010 int a[100005];
int main()
{
int n;
scanf("%d",&n);
for(int i=0; i<n; i++)
{
scanf("%d",&a[i]);
}
int flag=0;
int fl=0;
for(int i=1; i<n; i++)
{
if(flag==0)
{
if(a[i]>a[i-1])
continue;
else if(a[i]==a[i-1])
flag=1;
else if(a[i]<a[i-1])
flag=2;
}
else if(flag==1)
{
if(a[i]>a[i-1])
{
fl=1;
break;
}
else if(a[i]==a[i-1])
continue;
else if(a[i]<a[i-1])
flag=2;
}
else if(flag==2)
{
if(a[i]>=a[i-1])
{
fl=1;
break;
}
}
}
printf("%s\n",fl?"NO":"YES");
return 0;
}