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)或(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})__