cf914D(线段树+思维)

最新推荐文章于 2024-08-31 18:11:57 发布

qkoqhh

最新推荐文章于 2024-08-31 18:11:57 发布

阅读量611

点赞数

CC 4.0 BY-SA版权

分类专栏：线段树

本文链接：https://blog.youkuaiyun.com/qkoqhh/article/details/79127568

线段树专栏收录该内容

17 篇文章

订阅专栏

Bashanda喜欢猜测数组不同区间的最大公约数(GCD)，并希望改变最多一个元素使GCD接近他的猜测值。此问题通过构建区间树解决查询和更新操作，实现了高效的求解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

看到可以允许改一个数我就懵逼了..那我是不是得统计所有情况?感觉完全没思路..

然后汪聚聚提出一种思路..算出能被x整除的数的个数..然后就可以处理误差了...(真是好巧妙的思路蛙orz!!)

实际操作中只要处理不能被x整除的数,维护gcd,然后如果2个分支都不能整除就可以退出了,如果只有一个那么就要探到叶子,所以一次操作复杂度为logn

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<cmath>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define eps 1e-8
#define inf 1000000007
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define succ(x) (1<<x)
#define lowbit(x) (x&(-x))
#define sqr(x) ((x)*(x))
#define ls T[i<<1]
#define rs T[i<<1|1]
#define op T[i]
#define mid (x+y>>1)
#define NM 500005
#define nm 2000005
#define pi 3.141592653
using namespace std;
int read(){
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    return f*x;
}

int n,m,_y,_x,_t;
int gcd(int x,int y){return y==0?x:gcd(y,x%y);}

struct info{
	int s,size;
	info operator+(const info&o){
		info f;
		f.s=gcd(s,o.s);
		f.size=size+o.size;
		return f;
	}
}T[4*NM];

void build(int i,int x,int y){
	int t=x+y>>1;
	if(x==y){op.s=read();op.size=1;return;}
	build(i<<1,x,t);build(i<<1|1,t+1,y);
	op=ls+rs;
}

void mod(int i,int x,int y){
	int t=x+y>>1;
	if(x==y){op.s=_y;return;}
	if(_x<=t)mod(i<<1,x,t);
	else mod(i<<1|1,t+1,y);
	op=ls+rs;
}

int query(int i,int x,int y){
	int t=x+y>>1;
	if(_y<x||y<_x)return 0;
	if(_x<=x&&y<=_y){
		if(op.s%_t==0)return 0;
		if(x==y)return 1;
		if(ls.s%_t==0)return query(i<<1|1,t+1,y);
		if(rs.s%_t==0)return query(i<<1,x,t);
		return 2;
	}
	return query(i<<1,x,t)+query(i<<1|1,t+1,y);
}

int main(){
	//freopen("data.in","r",stdin);
	n=read();
	build(1,1,n);
	m=read();
	while(m--){
		_t=read();_x=read();_y=read();
		if(_t==1){
			_t=read();
			_t=query(1,1,n);
			puts(_t<2?"YES":"NO");
		}else mod(1,1,n);
	}
}

Bash likes playing with arrays. He has an array a₁, a₂, ... a_n of n integers. He likes to guess the greatest common divisor (gcd) of different segments of the array. Of course, sometimes the guess is not correct. However, Bash will be satisfied if his guess is almost correct.

Suppose he guesses that the gcd of the elements in the range [l, r] of a is x. He considers the guess to be almost correct if he can change at most one element in the segment such that the gcd of the segment is x after making the change. Note that when he guesses, he doesn't actually change the array — he just wonders if the gcd of the segment can be made x. Apart from this, he also sometimes makes changes to the array itself.

Since he can't figure it out himself, Bash wants you to tell him which of his guesses are almost correct. Formally, you have to process q queries of one of the following forms:

1 l r x — Bash guesses that the gcd of the range [l, r] is x. Report if this guess is almost correct.
2 i y — Bash sets a_i to y.

Note: The array is 1-indexed.

Input

The first line contains an integer n (1 ≤ n ≤ 5·10⁵) — the size of the array.

The second line contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹) — the elements of the array.

The third line contains an integer q (1 ≤ q ≤ 4·10⁵) — the number of queries.

The next q lines describe the queries and may have one of the following forms:

1 l r x (1 ≤ l ≤ r ≤ n, 1 ≤ x ≤ 10⁹).
2 i y (1 ≤ i ≤ n, 1 ≤ y ≤ 10⁹).

Guaranteed, that there is at least one query of first type.

Output

For each query of first type, output "YES" (without quotes) if Bash's guess is almost correct and "NO" (without quotes) otherwise.

Examples

Input

Output

YES
YES
NO

Input

Output

NO
YES
NO
YES

Note

In the first sample, the array initially is {2, 6, 3}.

For query 1, the first two numbers already have their gcd as 2.

For query 2, we can achieve a gcd of 3 by changing the first element of the array to 3. Note that the changes made during queries of type 1 are temporary and do not get reflected in the array.

After query 3, the array is now {9, 6, 3}.

For query 4, no matter which element you change, you cannot get the gcd of the range to be 2.