[Codeforces 920F] SUM and REPLACE

[题目链接]

         https://codeforces.com/contest/920/problem/F

[算法]

        显然 ， 一个10 ^ 6以内的数在经过最多6次操作后就会变为1或2 ， 这是因为一个数的因子个数是根号级别的

        用筛法预处理10 ^ 6以内每个数的因数个数 ， 用线段树维护即可 ， 修改时可以暴力地修改到叶子节点

        时间复杂度 : O(NlogN)

[代码]

        

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 3e5 + 10;
const int MAXP = 1e6 + 10;
typedef long long ll;
typedef long double ld;

int n , m;
int val[MAXN] , p[MAXP];

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;
}
struct Segment_Tree
{
        struct Node
        {
                int l ,r;
                ll cnt;
                bool flg;        
        }    a[MAXN << 2];
        inline void build(int index , int l , int r)
        {
                a[index].l = l , a[index].r = r;
                if (l == r)
                {
                        a[index].cnt = val[l];
                        a[index].flg = (a[index].cnt == 1 || a[index].cnt == 2);
                        return;
                }
                int mid = (l + r) >> 1;
                build(index << 1 , l , mid);
                build(index << 1 | 1 , mid + 1 , r);
                update(index);
        }
        inline bool update(int index)
        {
                a[index].flg = a[index << 1].flg & a[index << 1 | 1].flg;
                a[index].cnt = a[index << 1].cnt + a[index << 1 | 1].cnt;
        }
        inline void modify(int index , int l , int r)
        {
                int mid =(a[index].l + a[index].r) >> 1;
                if (a[index].flg) return;
                if (a[index].l == l && a[index].r == r)
                {
                        if (l == r) 
                        {
                                a[index].cnt = p[a[index].cnt];
                                a[index].flg = (a[index].cnt == 1 || a[index].cnt == 2);
                                return;
                        } else
                        {
                                modify(index << 1 , l , mid);
                                modify(index << 1 | 1 , mid + 1 , r);
                                update(index); 
                        }
                        return;
                }
                if (mid >= r) modify(index << 1 , l , r);
                else if (mid + 1 <= l) modify(index << 1 | 1 , l , r);
                else
                {
                        modify(index << 1 , l , mid);
                        modify(index << 1 | 1 , mid + 1 , r);
                }
                update(index);
        }
        inline ll query(int index , int l , int r)
        {
                if (a[index].l == l && a[index].r == r)
                        return a[index].cnt;
                else
                {
                        int mid = (a[index].l + a[index].r) >> 1;
                        if (mid >= r) return query(index << 1 , l , r);
                        else if (mid + 1 <= l) return query(index << 1 | 1 , l , r);
                        else return query(index << 1 , l , mid) + query(index << 1 | 1 , mid + 1 , r);
                }
        }
} SGT;

int main()
{
        
        read(n); read(m);
        for (int i = 1; i < MAXP; i++)
        {
                for (int j = i; j < MAXP; j += i)
                        ++p[j];        
        } 
        for (int i = 1; i <= n; i++) read(val[i]);
        SGT.build(1 , 1 , n);
        for (int i = 1; i <= m; i++)
        {
                int type;
                read(type);
                if (type == 1)
                {
                        int l , r;
                        read(l); read(r);
                        SGT.modify(1 , l , r);        
                }    else
                {
                        int l , r;
                        read(l); read(r);
                        printf("%lld
" , SGT.query(1 , l , r));
                }
        }
        
        return 0;
    
}