[Codeforces 914D] Bash and a Tough Math Puzzle

[题目链接]

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

[算法]

显然 ， 当一个区间[l , r]中为d倍数的数的个数 <= 1 , 答案为Yes , 否则为No

线段树简单维护即可 ， 详见代码 , 时间复杂度 : O(NlogN ^ 2)

[代码]

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

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

struct Segment_Tree
{
    struct Node
    {
        int l , r;
        int value;
    } a[MAXN << ];
    inline int gcd(int x , int y)
    {
        if (y == ) return x;
        else return gcd(y , x % y);
    }
    inline void update(int x)
    {
        a[x].value = gcd(a[x << ].value , a[x <<  | ].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) >> ;
        build(index <<  , l , mid);
        build(index <<  |  , mid +  , 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) >> ;
            if (mid >= x) modify(index <<  , x , y);
            else modify(index <<  |  , x , y);
            update(index);
        }
    }
    inline void getans(int index , int l , int r , int d)
    {
        if (cnt > ) return;
        int mid = (a[index].l + a[index].r) >> ;
        if (a[index].l == l && a[index].r == r)
        {
            if (l == r)
            {
                if (a[index].value % d)
                    ++cnt;
                return;
            }
            if ((a[index << ].value % d) && (a[index <<  | ].value % d))
            {
                cnt += ;
                return;
            } else if (a[index << ].value % d) getans(index <<  , l , mid , d);
            else if (a[index <<  | ].value % d) getans(index <<  |  , mid +  , r , d);
        } else
        {
            if (mid >= r) getans(index <<  , l , r , d);
            else if (mid +  <= l) getans(index <<  |  , l , r , d);
            else
            {
                getans(index <<  , l , mid , d);
                getans(index <<  |  , mid +  , 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 = ; x = ;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << ) + (x << ) + c - '';
    x *= f;
}

int main()
{

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

    return ;
}