Bash likes playing with arrays. He has an array a1, a2, ... an 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 xafter 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 ai to y.
Note: The array is 1-indexed.
The first line contains an integer n (1 ≤ n ≤ 5·105) — the size of the array.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 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.
For each query of first type, output "YES" (without quotes) if Bash's guess is almost correct and "NO"(without quotes) otherwise.
3 2 6 3 4 1 1 2 2 1 1 3 3 2 1 9 1 1 3 2
YES YES NO
5 1 2 3 4 5 6 1 1 4 2 2 3 6 1 1 4 2 1 1 5 2 2 5 10 1 1 5 2
NO YES NO YES
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.
題意：給出n個數，q個操作
操作一：求[l,r]的gcd是否爲x，是輸出YES,否輸出NO(如果可以改變[l,r]中的一個數使得gcd爲x，依然輸出YES)
操作二：將x位置的數改爲y
分析：操作二很好搞直接單點更新就行了，那麼只用考慮操作一；要使區間的gcd爲x，這個區間內的所有數一定是x的倍數，因此只需要找區間內不是x倍數的數字有多少個就行了，如果>1個則是NO
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=5e5+500;
int t[N<<2];
void build(int l,int r,int rt)
{
if(l==r)
{
scanf("%d",&t[rt]);
return;
}
int m=(l+r)>>1;
build(l,m,rt<<1);
build(m+1,r,rt<<1|1);
t[rt]=__gcd(t[rt<<1],t[rt<<1|1]);
}
void update(int pos,int l,int r,int rt,int num)
{
if(l==r)
{
t[rt]=num;
return;
}
int m=(l+r)>>1;
if(pos<=m)
update(pos,l,m,rt<<1,num);
else
update(pos,m+1,r,rt<<1|1,num);
t[rt]=__gcd(t[rt<<1],t[rt<<1|1]);
}
int query(int ql,int qr,int l,int r,int rt,int num)
{
if(l==r)return 1;
int m=(l+r)>>1,ans=0;
if(ql<=m&&t[rt<<1]%num!=0)
ans+=query(ql,qr,l,m,rt<<1,num);
if(qr>m&&t[rt<<1|1]%num!=0&&ans<=1)//這裏<=1剪個枝可以節省很多時間
ans+=query(ql,qr,m+1,r,rt<<1|1,num);
return ans;
}
int main()
{
int n,q,op,l,r,x,y;
scanf("%d",&n);
build(1,n,1);
scanf("%d",&q);
while(q--)
{
scanf("%d",&op);
if(op==1)
{
scanf("%d%d%d",&l,&r,&x);
if(query(l,r,1,n,1,x)<=1)
puts("YES");
else puts("NO");
}
else
{
scanf("%d%d",&x,&y);
update(x,1,n,1,y);
}
}
return 0;
}