[題目連結]

         https://codeforces.com/contest/914/problem/D

[演算法]

         顯然 ， 當一個區間[l , r]中為d倍數的數的個數 <= 1 , 答案為Yes , 否則為No

         線段樹簡單維護即可 ， 詳見程式碼 , 時間複雜度 : O(NlogN ^ 2)

[程式碼]

#include<bits/stdc++.h>
using
 
 namespace std;
const int MAXN = 5e5 + 10;

int n , m;
int val[MAXN];
int cnt;

struct Segment_Tree
{
    struct Node
    {
        int l , r;
        int value;    
    } a[MAXN << 2];
    inline int gcd(int x , int y)
    {
        if (y == 0) return x;
        else return gcd(y , x % y);
    }
    inline 
 
void update(int x)
    {
        a[x].value = gcd(a[x << 1].value , a[x << 1 | 1].value);
    }
    inline void build(int index , int l , int r)
    {
        a[index].l = l , a[index].r = r;
        if (l == r)
        {
            a[index].value = val[l];
            return;
        }
        
 
int mid = (l + r) >> 1;
        build(index << 1 , l , mid);
        build(index << 1 | 1 , mid + 1 , r);
        update(index);
    }
    inline void modify(int index , int x , int y)
    {
        if (a[index].l == a[index].r)
        {
            a[index].value = y;
            return;
        } else
        {
            int mid = (a[index].l + a[index].r) >> 1;
            if (mid >= x) modify(index << 1 , x , y);
            else modify(index << 1 | 1 , x , y);
            update(index);
        }
    }
    inline void getans(int index , int l , int r , int d)
    {
        if (cnt > 1) return;
        int mid = (a[index].l + a[index].r) >> 1;
        if (a[index].l == l && a[index].r == r)
        {
            if (l == r) 
            {
                if (a[index].value % d) 
                    ++cnt;
                return;
            }
            if ((a[index << 1].value % d) && (a[index << 1 | 1].value % d)) 
            {
                cnt += 2;
                return;
            } else if (a[index << 1].value % d) getans(index << 1 , l , mid , d);
            else if (a[index << 1 | 1].value % d) getans(index << 1 | 1 , mid + 1 , r , d);
        } else
        {
            if (mid >= r) getans(index << 1 , l , r , d);
            else if (mid + 1 <= l) getans(index << 1 | 1 , l , r , d);
            else 
            {
                getans(index << 1 , l , mid , d);
                getans(index << 1 | 1 , mid + 1 , r , d);
            }
        }
    }
} SGT;

template <typename T> inline void chkmax(T &x , T y) { x = max(x , y); }
template <typename T> inline void chkmin(T &x , T y) { x = min(x , y); }
template <typename T> inline void read(T &x) 
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
    x *= f;        
}

int main()
{
    
    read(n);
    for (int i = 1; i <= n; i++) read(val[i]);
    SGT.build(1 , 1 , n);
    read(m);
    while (m--)
    {
        int type;
        read(type);
        if (type == 1)
        {
            int l , r , x;
            read(l); read(r); read(x);
            cnt = 0;
            SGT.getans(1 , l , r , x);
            if (cnt > 1) printf("NO\n");
            else printf("YES\n");    
        } else
        {
            int x , y;
            read(x); read(y);
            SGT.modify(1 , x , y);
        }
    }
    
    return 0;
}