Codeforces Round #835 (Div. 4) - D. Challenging Valleys

D. Challenging Valleys

枚举所有相等的连续段，判断符合条件的是否只有一段即可。

#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for(int i = a;i<n;i++)
#define per(i,a,n) for(int i = n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define eb emplace_back
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define yes cout<<"YES"<<'\n';
#define no cout<<"NO"<<'\n';
#define endl '\n';
typedef vector<int> VI;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef double db;
mt19937 mrand(random_device{}());
const ll mod=1000000007;
int rnd(int x) {return mrand() % x;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;};
ll lcm(int a,int b){return a*b/gcd(a,b);}const int N=2e5+10;
int T;
int n;
int a[N];void solve(){cin>>n;int cnt=0;rep(i,0,n) cin>>a[i];int l=0,r=0;while(l<n&&r<n){while(r<n-1&&a[r]==a[r+1]) r++;if((l==0||a[l-1]>a[l])&&(r==n-1||a[r]<a[r+1])){cnt++;}l=r+1,r=l;}puts(cnt==1?"YES":"NO");
}int main(){ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);cin>>T;while(T--){solve();}return 0;
}

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