【WOJ4350】ldx的SGT水题2

水题 gcd可合并

#include<bits/stdc++.h>
const int N=200005;
using namespace std;
template<class T>
inline void read(T &x)
{
	x=0;
	static char ch=getchar();
	while(!isdigit(ch))	ch=getchar();
	while(isdigit(ch)) x=x*10+ch-'0',ch=getchar();
}
struct Tree
{
	int l,r,gcd;
}tree[4*N];
int n,val[N];
inline void pushup(int now)
{
	tree[now].gcd=__gcd(tree[2*now].gcd,tree[2*now+1].gcd);
}
inline void build(int now,int l,int r)
{
	tree[now].l=l; tree[now].r=r;
	if(l==r)
	{
		tree[now].gcd=val[l];
		return;
	}
	int m=(l+r)>>1;
	build(2*now,l,m);
	build(2*now+1,m+1,r);
	pushup(now);
}
void update(int now,int k,int v)
{
	if(tree[now].l==k&&tree[now].r==k)
	{
		 tree[now].gcd=v;
		 return;
	}
	int m=(tree[now].l+tree[now].r)>>1;
	if(k<=m)	update(2*now,k,v);
	else update(2*now+1,k,v);
	pushup(now);
}
int query(int now,int l,int r)
{
	if(l<=tree[now].l&&tree[now].r<=r) return tree[now].gcd;
	int m=(tree[now].l+tree[now].r)>>1,ans=0;
	if(l<=m) ans=__gcd(ans,query(2*now,l,r));
	if(r>m)	ans=__gcd(ans,query(2*now+1,l,r));
	return ans;
}
int main()
{
	cin>>n;
	for(int i=1;i<=n;i++)	read(val[i]);
	build(1,1,n);
	int T;
	cin>>T;
	while(T--)
	{
		int l,r,x;
		read(x); read(l); read(r);
		if(x==0)	cout<<query(1,l,r)<<'\n';
		else update(1,l,r);
	}
	return 0;
}