link

思路：

考慮分塊，每個塊都用deque維護包含的元素，方便刪除和新增。

程式碼：

// Problem: CF455D Serega and Fun
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/CF455D
// Memory Limit: 250 MB
// Time Limit: 4000 ms
// 
// Powered by CP Editor (https://cpeditor.org)
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll>PLL;
typedef pair<int, int>PII;
typedef pair<double, double>PDD;
#define I_int ll
inline ll read()
{
    ll x = 0, f = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
        if(ch == '-')f = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9')
    {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}
  
inline void out(ll x){
    if (x < 0) x = ~x + 1, putchar('-');
    if (x > 9) out(x / 10);
    putchar(x % 10 + '0');
}
  
inline void write(ll x){
    if (x < 0) x = ~x + 1, putchar('-');
    if (x > 9) write(x / 10);
    putchar(x % 10 + '0');
}
  
#define read read()
#define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define multiCase int T;cin>>T;for(int t=1;t<=T;t++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i<(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define perr(i,a,b) for(int i=(a);i>(b);i--)
ll ksm(ll a, ll b,ll p)
{
    ll res = 1;
    while(b)
    {
        if(b & 1)res = res * a%p ;
        a = a * a %p;
        b >>= 1;
    }
    return res;
}

const int inf = 0x3f3f3f3f;
const int maxn=1e5+7,maxm=210000;

int n,c[360][maxn];
deque<int>a[maxn];
int num,block,belong[maxn],l[maxn],r[maxn];
int ans=0;

void cul(int &x){
	x=(x+ans-1)%n+1;
}

void update(int l,int r){
	int bl=l/block,br=r/block;
	if(bl==br){
		r=r%block-1;
		int x=a[br][r];
		a[br].erase(a[br].begin()+r);
		a[bl].insert(a[bl].begin()+l%block,x);
	}
	else{
		int x;
		for(int i=bl;i<br;){
			x=a[i].back();a[i].pop_back();c[i][x]--;
			i++;
			a[i].push_front(x);c[i][x]++;
		}
		r%=block;x=a[br][r];
		a[br].erase(a[br].begin()+r);c[br][x]--;
		a[bl].insert(a[bl].begin()+l%block,x);c[bl][x]++;
	}
}

int query(int l,int r,int k){
	int an=0;
	int bl=l/block,br=r/block;
	if(bl==br){
		for(int i=l%block;i<r%block;i++)
			if(a[bl][i]==k) an++;
	}
	else{
		for(int i=bl+1;i<br;i++)
			an+=c[i][k];
		for(int i=l%block;i<a[i].size();i++)
			if(a[bl][i]==k) an++;
		for(int i=0;i<r%block;i++)
			if(a[br][i]==k) an++;
	}
	return an;
	
}

int main(){
	n=read;
	block=int(sqrt(n));
	rep(i,0,n-1){
		int x=read;
		a[i/block].push_back(x);
		c[i/block][x]++;
	}
	int q=read;
	while(q--){
		int op=read,l=read,r=read;
		cul(l);cul(r);
		if(l>r) swap(l,r);
		l--;
		if(op==1) update(l,r);
		else{
			int k=read;cul(k);
			ans=query(l,r,k);
			printf("%d\n",ans);
		}
	}
	
	
	
	
	
	return 0;
}