Codeforces Round #424 (Div. 2, rated, based on VK Cup Finals)
昨晚的没来得及打,最近错过好几场CF了,这场应该不算太难
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).
这个A可能比B还要难一点,给你一个序列,满足左侧严格递增,中间相等,右侧严格递减,左右也可以为空,你判定下
#include <bits/stdc++.h> using namespace std; int main() { int n; int a[105]; cin>>n; for(int i=1; i<=n; i++) cin>>a[i]; int f1=1; a[n+1]=a[0]=1<<30; int f=2; while(a[f]>a[f-1]) f++; while(a[f]==a[f-1]) f++; while(a[f]<a[f-1]) f++; if(f<=n) cout<<"NO"<<endl; else cout<<"YES"<<endl; return 0; }
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