Interval GCD (see the interval gcd for modification of the line segment tree)

D. Interval_GCD

Blue book title Interval GCD
title meaning: Given n numbers Ai, M operations (N: 5e5, M: 5e5) have two operations:

• C lrd adds d to A[l,r]
• Q lr asks the greatest common divisor of A[l,r]

思路：
$\gcd (a_1,a_2,a_3,a_4...,a_n)=\gcd (a_1,a_2-a_1,a_3-a_2,a_4-a_3...a_n-a_{n-1})$
所以: $\gcd (a_l,a_{l+1},a_{l+2}...a_r)=gcd(a_l,a_{l+1}-a_l,a_{l+2}-a_{l+1}...a_r-a_{r-1})$
The sequence maintained in this way is converted from the original sequence A to the difference sequence B of A, Now all that remains is for $a_l$The
advantage of maintenance is:
for each operation C (interval addition), only a single point of modification of the ll of the B sequence$l$ and$r+1$ position (llAdd d to$l$ ;$r+1$ minus d)

Method : Establish a line segment tree to maintain differential sequence B (in order to calculate the gcd of differential sequence B $gcd$ )
Establish the grooming number maintenance series A (in order to calculate$a_l$）

struct SegmentTree{
    
    
    int l,r;
    LL dat;//计算gcd
    #define l(x) tree[x].l
    #define r(x) tree[x].r
    #define dat(x) tree[x].dat
}tree[maxn*4];
LL a[maxn],b[maxn],c[maxn];
int n,m;

void build(int p,int l,int r){
    
    
    l(p)=l,r(p)=r;
    if(l==r){
    
    dat(p)=b[l];return;}
    int mid=(l+r)/2;
    build(p*2,l,mid);
    build(p*2+1,mid+1,r);
    dat(p)=lgcd(dat(p*2),dat(p*2+1));
}

/*void lazy(int p){
    if(add(p)){
        sum(p*2)+=add(p)*(r(p*2)-l(p*2)+1);
        sum(p*2+1)+=add(p)*(r(p*2+1)-l(p*2+1)+1);
        add(p*2)+=add(p);
        add(p*2+1)+=add(p);
        add(p)=0;
    }
}*/

void change(int p,int x,LL z){
    
    
    if(l(p)==r(p)){
    
    dat(p)+=z;return;}
    int mid=(l(p)+r(p))/2;
    if(x<=mid) change(p*2,x,z);
    else change(p*2+1,x,z);
    dat(p)=lgcd(dat(p*2),dat(p*2+1));
}

LL ask(int p,int l,int r){
    
    
    if(l<=l(p)&&r>=r(p)) return abs(dat(p));
    int mid=(l(p)+r(p))/2;
    LL ans=0ll;
    if(l<=mid)ans=lgcd(ans,ask(p*2,l,r));
    if(r>mid)ans=lgcd(ans,ask(2*p+1,l,r));
    return abs(ans);
}

void updata(int i,LL k){
    
    //在i位置上加上k
    while(i<=n){
    
    
        c[i]+=k;
        i+=i&(-i);
    }
}///区间[p,q]加c：updata(p,c);updata(q+1,-c);

LL getsum(int i){
    
    //求i的前缀和(1~i)
    LL res=0;
    while(i){
    
    
        res+=c[i];
        i-=i&(-i);
    }
    return res;
}///区间[p,q]:getsum(q)-getsum(p-1)

int main(){
    
    
    cin>>n>>m;
    memset(c,0,sizeof c);
    for(int i=1;i<=n;i++){
    
    cin>>a[i];b[i]=a[i]-a[i-1];}
    build(1,1,n);
    int x,y;
    LL z;
    char op;
    while(m--){
    
    
        cin>>op>>x>>y;
        if(op=='C'){
    
    
            cin>>z;
            updata(x,z);//维护数组A
            change(1,x,z);//维护数组B
            if(y+1<=n){
    
    change(1,y+1,-1ll*z);updata(y+1,-1l*z);}
        }else{
    
    cout<<lgcd(a[x]+getsum(x),ask(1,x+1,y))<<endl;}
    }
}