题意

给定序列$a_n$，每次将$[L,R]$区间内的数$a_i$替换为$d(a_i)$，或者询问区间和

---

这题和区间开方有相同的操作

对于$a_i \in (1,10^6)$，$10$次$d(a_i)$以内肯定可以最终化为$1$或者$2$，所以线段树记录区间最大值和区间和，$Max\le2$就返回，单点暴力更新，最后线性筛预处理出$d$

时间复杂度$O(m\log n)$

代码

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

const int N = 300005;

const int M = 1000005;

int a[N];

int lch[N << 2], rch[N << 2];

LL sum[N << 2], Max[N << 2];

LL prime[M], e[M], d[M];

int check[M], cnt;

void seive() {

    memset(check, 0, sizeof(check)); cnt = 0; d[1] = 1; e[1] = 1;

    for(int i = 2; i < M; ++i) {

        if(!check[i]) {prime[cnt++] = i; e[i] = 1; d[i] = 2;}

        for(int j = 0; j < cnt; ++j) {

            if(1LL * i * prime[j] >= M) break;

            check[i * prime[j]] = 1;

            if(i % prime[j] == 0) {

                e[i * prime[j]] = e[i] + 1;

                d[i * prime[j]] = d[i] / (e[i] + 1) * (e[i] + 2);

                break;

            }else {

                d[i * prime[j]] = d[i] * 2; e[i * prime[j]] = 1;

            }

        }

    }

}

inline void pushup(int x) {

    sum[x] = sum[x << 1] + sum[x << 1 | 1];

    Max[x] = max(Max[x << 1], Max[x << 1 | 1]);

}

void build(int x, int l, int r) {

    lch[x] = l; rch[x] = r; sum[x] = Max[x] = 0;

    if(l == r) {

        sum[x] = Max[x] = a[l]; return;

    }

    int mid = (l + r) / 2;

    build(x << 1, l, mid); build(x << 1 | 1, mid + 1, r);

    pushup(x);

}

void update(int x, int l, int r) {

    if(Max[x] <= 2) return;

    if(lch[x] == rch[x]) {

        sum[x] = Max[x] = d[sum[x]]; return;

    }

    int mid = (lch[x] + rch[x]) / 2;

    if(r <= mid) update(x << 1, l, r);

    else if(l > mid) update(x << 1 | 1, l, r);

    else update(x << 1, l, mid), update(x << 1 | 1, mid + 1, r);

    pushup(x);

}

LL query(int x, int l, int r) {

    if(l == lch[x] && rch[x] == r) return sum[x];

    int mid = (lch[x] + rch[x]) / 2;

    if(r <= mid) return query(x << 1, l, r);

    else if(l > mid) return query(x << 1 | 1, l, r);

    else return query(x << 1, l, mid) + query(x << 1 | 1, mid + 1, r);

}

int n, m, t, l, r;

int main() {

    seive();

    // for (int i = 1; i < M; i++)

    //     for (int j = i; j < M; j += i) d[j]++;

    scanf("%d%d", &n, &m);

    for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);

    build(1, 1, n);

    for(int i = 1; i <= m; ++i) {

        scanf("%d%d%d", &t, &l, &r);

        if(t == 1) update(1, l, r);

        else printf("%I64d\n", query(1, l, r));

    }

    return 0;

}