题解 CF920F 【SUM and REPLACE】

题目描述

维护一个长度为(n)序列,支持对一段区间将其所有元素换成其正约数个数,支持区间求和,(m)组询问

对于(1le n,mle 3 imes 10^5),对于序列元素(1le a_{i}le 10^6)

解题思路

一看维护区间修改区间求和想到线段树,区间求和很好办,但是第一个貌似很棘手,考虑正约数个数( au)是一个积性函数,数据范围(1le a_{i}le 10^6),因此我们可以用线性筛将最大元素内的所有( au)值筛出来然后修改时候直接修改即可。

这样暴力修改部分点会(T),但是卡卡常数貌似可过。

考虑我们什么操作是多余的,对于一段区间,若是全为(1)(2),我们在修改的时候可以直接跳过,因此再维护一个区间最大值,修改的时候若最大值小于(3)直接忽略就可以

(egin{aligned}mathcal{Code}end{aligned})

#include<bits/stdc++.h>
    
#define LL long long
    
#define _ 0
    
using namespace std;
    
/*Grievous Lady*/
    
const int BUF_SIZE = 1 << 12;
    
char buf[BUF_SIZE] , *buf_s = buf , *buf_t = buf + 1;
    
#define PTR_NEXT() 
{ 
    buf_s ++; 
    if(buf_s == buf_t) 
    { 
        buf_s = buf; 
        buf_t = buf + fread(buf , 1 , BUF_SIZE , stdin); 
    } 
}
    
template <typename _m_> void mian(_m_ & _n_){
    int _x_ = 0 , _nega_ = 0;
    while(*buf_s != '-' && !isdigit(*buf_s)) PTR_NEXT(); if(*buf_s == '-'){_nega_ = 1; PTR_NEXT();}
    while(isdigit(*buf_s)){_x_ = _x_ * 10 + *buf_s - '0'; PTR_NEXT();} if(_nega_) _x_ = -_x_; (_n_) = (_x_);
}
    
const int kato = 1e6 + 10;

int prime[kato] , tau[kato] , num[kato];

LL cnt , max_;

bool ispri[kato];

inline void pri(int n){
    for(int i = 2;i <= n;i ++){
        if(!ispri[i]){
            prime[++ cnt] = i;
            num[i] = 1;
            tau[i] = 2;
        }
        for(int j = 1;j <= cnt && i * prime[j] <= n;j ++){
            ispri[i * prime[j]] = 1;
            if(i % prime[j] == 0){
                num[i * prime[j]] = num[i] + 1;
                tau[i * prime[j]] = tau[i] / num[i * prime[j]] * (num[i * prime[j]] + 1);
                break;
            }
            num[i * prime[j]] = 1;
            tau[i * prime[j]] = tau[i] * tau[prime[j]];
        }
    }
}

struct tree{
    protected:

        struct node{
            node *ch[2];
            int l , r;
            LL sum , tag;
            node(int l = 0 , int r = 0 , LL sum = 0 , LL tag = 0): l(l) , r(r) , sum(sum) , tag(tag){
                ch[0] = ch[1] = NULL;
            }
            inline int mid(){
                return (l + r) >> 1;
            }
            inline void up(){
                sum = ch[0] -> sum + ch[1] -> sum;
                tag = max(ch[0] -> tag , ch[1] -> tag);
            }
        }*root;

        inline void build(node *&o , int l , int r){
            o = new node(l , r);
            if(l == r){
                mian(o -> sum);
                o -> tag = o -> sum;
                max_ = max(o -> sum , max_);
                return;
            }
            build(o -> ch[0] , l , o -> mid()); build(o -> ch[1] , o -> mid() + 1 , r);
            o -> up();
        }

        inline void Modify(node *o , int l , int r){
            if(o -> tag < 3) return;
            if(o -> l == o -> r){
                o -> tag = o -> sum = tau[o -> sum];
                return;
            }
            if(l <= o -> mid()){
                Modify(o -> ch[0] , l , r);
            }
            if(r > o -> mid()){
                Modify(o -> ch[1] , l , r);
            }
            o -> up();
        }

        inline LL ask(node *o , int l , int r){
            // if(o -> l > r || o -> r < l) return 0;
            if(l <= o -> l && o -> r <= r){
                return o -> sum;
            }
            LL res = 0;
            if(l <= o -> mid()){
                res += ask(o -> ch[0] , l , r);
            }
            if(r > o -> mid()){
                res += ask(o -> ch[1] , l , r);
            }
            return res;
        }
        
    public:

        inline void build(int n){
            build(root , 1 , n);
        }

        inline void Modify(int l , int r){
            Modify(root , l , r);
        }

        inline LL ask(int l , int r){
            return ask(root , l , r);
        }
}yuni;

int n , m , opt , l , r;

inline int Ame_(){
    mian(n) , mian(m);
    yuni.build(n);
    tau[1] = 1;
    pri(max_);
    for(; m --> 0 ;){
        mian(opt) , mian(l) , mian(r);
        if(opt == 1) yuni.Modify(l , r);
        if(opt == 2) printf("%lld
" , yuni.ask(l , r));
    }
    return ~~(0^_^0);
}
    
int Ame__ = Ame_();
    
int main(){;}

-----(egin{aligned}mathcal{Ame}end{aligned})__

原文地址:https://www.cnblogs.com/Ame-sora/p/13747827.html