区间gcd

Description

区间加减、区间 $gcd$

Solution

因为 $gcd(a,b)=gcd(a,b-a)$，
所以可以差分， $gcd(a_l,a_{l+1},...,a_r)=gcd(a_l,a_{l+1}-a_l,...,a_r-a_{r-1})$

Code

树状数组维护 $a[i]$，线段树维护 $a[i]-a[i-1]$

#include<bits/stdc++.h>
using namespace std;
#define mid ((l+r)>>1)
const int N=100002;
int s[N],a[N],b[N],g[N<<2],n,m,i,l,r,op,x;
inline char gc(){
	static char buf[100000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline int rd(){
	int x=0,fl=1;char ch=gc();
	for(;ch<48||ch>57;ch=gc())if(ch=='-')fl=-1;
	for(;48<=ch&&ch<=57;ch=gc())x=(x<<3)+(x<<1)+(ch^48);
	return x*fl;
}
inline void wri(int a){if(a<0)a=-a,putchar('-');if(a>=10)wri(a/10);putchar(a%10|48);}
inline void wln(int a){wri(a),puts("");}
void add(int x,int y){
	for (;x<=n;x+=x&-x) s[x]+=y;
}
int ask(int x){
	int ans=0;
	for (;x;x^=x&-x) ans+=s[x];
	return ans;
}
void build(int t,int l,int r){
	if (l==r){
		g[t]=b[l];
		return;
	}
	build(t<<1,l,mid),build(t<<1|1,mid+1,r);
	g[t]=__gcd(g[t<<1],g[t<<1|1]);
}
void update(int t,int l,int r,int x,int y){
	if (l==r){
		g[t]+=y;
		return;
	}
	if (x<=mid) update(t<<1,l,mid,x,y);
	else update(t<<1|1,mid+1,r,x,y);
	g[t]=__gcd(g[t<<1],g[t<<1|1]);
}
int query(int t,int l,int r,int x,int y){
	if (x<=l && r<=y) return g[t];
	int ans=0;
	if (x<=mid) ans=query(t<<1,l,mid,x,y);
	if (mid<y) ans=__gcd(ans,query(t<<1|1,mid+1,r,x,y));
	return ans;
}
int main(){
	n=rd(),m=rd();
	for (i=1;i<=n;i++) a[i]=rd(),b[i]=a[i]-a[i-1],add(i,a[i]),add(i+1,-a[i]);
	build(1,1,n);
	for (;m--;){
		op=rd(),l=rd(),r=rd();
		if (op==1){
			x=rd(),add(l,x),update(1,1,n,l,x);
			if (r<n) add(r+1,-x),update(1,1,n,r+1,-x);
		}
		else wln(abs(__gcd(query(1,1,n,l+1,r),ask(l))));
	}
}