输入时因为只用计算差，可以只用两个变量迭代，计算有不有趣也可以用桶排搞定

int n,x,y;
bool a[3001];
int main(){
	while(cin>>n){
		memset(a,0,sizeof(a));
        cin>>x;
		for(int i=2;i<=n;i++){
            cin>>y;
			int cha=abs(y-x);
            if(cha<n)a[cha]=1;
            x=y;
		}
		int cnt=0;
		for(int i=1;i<=n-1;i++)if(a[i])cnt++;
		if(cnt==n-1)cout<<"Jolly\n";
		else cout<<"Not jolly\n";
	}
	return 0;
}

/*****************************************
备注：
******************************************/
#include <queue>
#include <math.h>
#include <stack>
#include <stdio.h>
#include <iostream>
#include <vector>
#include <iomanip>
#include <string.h>
#include <algorithm>
using namespace std;
#define LL long long
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
int a[N];
int main()
{
	int n ;
	while(cin >> n)
	{
		memset(a,0,sizeof(a));//string.h  初始化数组
		int pre;
		cin >> pre;
		int now;
		for(int i = 1 ; i < n ; i++)
		{
			cin >> now;
			int s = now - pre;
			if(s < 0)
				s = s * -1;
			a[s] = 1;
			pre = now;
		}
		int flag = 1;
		for(int i = 1 ; i < n ; i++)
		{
			if(a[i] == 0)
			{
				flag=  0;
				break;
			}
		}
		if(flag == 1)
			cout << "Jolly\n";
		else 
			cout << "Not jolly\n";
	}
	return 0;
}

#include<bits/stdc++.h>
using namespace std;
const int N=3e3+10;
int n,x,a[N];
bool v[N];
int main(){
	while(cin>>n){
		int ans=0;
		cin>>a[1];
		for(int i=2;i<=n;i++){
			cin>>a[i];
			if(abs(a[i]-a[i-1])<n&&v[abs(a[i]-a[i-1])]==0){
				v[abs(a[i]-a[i-1])]=1;
				ans++;
			}
		}
		if(ans==n-1){
			cout<<"Jolly"<<endl;
		}else{
			cout<<"Not jolly"<<endl;
		}
		memset(v,0,sizeof(v));
	}
	return 0;
}

#include<bits/stdc++.h>
using namespace std;
const int N =3e3+10;
int n,a[N];
bool v[N];
int main(){
	while(cin>>n){
		memset(v,0,sizeof(v));
		int sum=0;//计算有多少个标记的数
        cin>>a[1];
		for(int i=2;i<=n;i++)
		{
            cin>>a[i];
            if(abs(a[i]-a[i-1])<n&&v[abs(a[i]-a[i-1])]==0)
			{
				v[abs(a[i]-a[i-1])]=1;//标记
				sum++;
			}
		}
		if(sum==n-1)
			cout<<"Jolly"<<endl;
		else 
			cout<<"Not jolly"<<endl;
	}
	return 0;
}

#include <bits/stdc++.h>
using namespace std;
const int N = 3e3 + 10;
int n,a[N];
bool v[N];
int main(){
	while(cin >> n){
		memset(v,0,sizeof(v));
		int sum = 0;
		cin >> a[1];
		for(int i = 2;i <= n;i++){
			cin >> a[i];
			if(abs(a[i] - a[i - 1]) < n && v[abs(a[i] - a[i - 1])] == 0){
				v[abs(a[i] - a[i - 1])] = 1;
				sum++;
			}
		}
		if(sum == n - 1){
			cout << "Jolly" << endl;
		}else{
			cout << "Not jolly" << endl;
		}
	}
	return 0;
}