C. Shinju and the Lost Permutation

// Problem: C. Shinju and the Lost Permutation
// Contest: Codeforces - Codeforces Round #779 (Div. 2)
// URL: https://codeforces.com/contest/1658/problem/C
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 2022-03-27 22:59:43
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;   
typedef vector<ll> Vll;               
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;                        

const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const double pi  = acos(-1.0);

inline ll qmi (ll a, ll b) {
	ll ans = 1;
	while (b) {
		if (b & 1) ans = ans * a%mod;
		a = a * a %mod;
		b >>= 1;
	}
	return ans;
}
inline int read () {
	int x = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
	return f?-x:x;
}
template<typename T> void print(T x) {
	if (x < 0) putchar('-'), x = -x;
	if (x >= 10) print(x/10);
	putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
	return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
	return (a + b) %mod;
}
inline ll inv (ll a) {
	return qmi(a, mod - 2);
}
void solve() {
	int n;
	cin >> n;
	vector<int> a(n);
	int c =0;
	for (int i = 0; i < n; i ++) {
		cin >> a[i];
		if (a[i] == 1) {
			c ++;
		}
	}
	if (c > 1|| c == 0) {
		puts("NO");
		return;
	}
	for (int i = 0; i < n - 1; i ++)
		if (a[i + 1] - a[i] > 1) {
			puts("NO");
			return;
		}
	if (a[0] - a[n - 1] > 1) {
		puts("NO");
		return;
	}
	puts("YES");
}
int main () {
	// ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
    int t;
    t =1;
    cin >> t;
    while (t --) solve();
    return 0;
}

思维就是一个算法中的小细节
这题可以用单调栈的思想来解答