F. SUM and REPLACE
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Let D(x) be the number of positive divisors of a positive integer x. For example, D(2) = 2 (2 is divisible by 1 and 2), D(6) = 4 (6 is divisible by 1, 2, 3 and 6).
You are given an array a of n integers. You have to process two types of queries:
REPLACE l r — for every replace ai with D(ai);
SUM l r — calculate .
Print the answer for each SUM query.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 3·105) — the number of elements in the array and the number of queries to process, respectively.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 106) — the elements of the array.
Then m lines follow, each containing 3 integers ti, li, ri denoting i-th query. If ti = 1, then i-th query is REPLACE li ri, otherwise it's SUM li ri (1 ≤ ti ≤ 2, 1 ≤ li ≤ ri ≤ n).
There is at least one SUM query.
Output
For each SUM query print the answer to it.
Example
inputCopy
7 6
6 4 1 10 3 2 4
2 1 7
2 4 5
1 3 5
2 4 4
1 5 7
2 1 7
outputCopy
30
13
4
22
https://codeforces.com/contest/920/problem/F
题意：
给你一个含有n个数的数组，和m个操作
操作1：将l~r中每一个数(a[i])变成 (d(a[i]))
其中$ d(x)$ 是约数个数函数。
操作2： 求l~r的a[i] 的sum和。
思路：
$ d(x)$ 约数个数函数可以利用线性筛预处理处理。
又因为 (d(2)=2) 和 (d(1)=1) 操作1对a[i]等于1或者2没有影响。
那么我们可以对一个区间中全都是1或者2不更新操作。
同时 (d(x)) 是收敛函数， 在1e6 的范围内，最多不超过5次改变就会收敛到1或2.
所以更新操作可以暴力解决，
同时用线段树维护即可。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), ' ', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {a %= MOD; if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("
");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("
");}}}
inline void getInt(int *p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
// d(n)表示n的约数个数和
// prime[i]表示第i个质数
//num[i]表示i的最小质因子出现次数
int sshu[maxn];
int N = maxn;
int num[maxn];
int d[maxn];
bool no[maxn];
int tot;
void prepare()
{
d[1] = 1; num[1] = 1;
for (int i = 2; i < N; i++) {
if (!no[i]) {
sshu[++tot] = i;
d[i] = 2; num[i] = 1;
}
for (int j = 1; j <= tot && sshu[j]*i < N; j++) {
int v = sshu[j] * i;
no[v] = 1;
if (i % sshu[j] == 0) {
num[v] = num[i] + 1;
d[v] = d[i] / num[v] * (num[v] + 1);
break;
}
d[v] = d[i] << 1; num[v] = 1;
}
}
//for (int i=1;i<=10;i++) printf("%d
",d[i]);
}
int a[maxn];
struct node {
int l, r;
int laze;
bool isall;
ll num;
} segment_tree[maxn << 2];
void pushup(int rt)
{
segment_tree[rt].num = segment_tree[rt << 1].num + segment_tree[rt << 1 | 1].num;
segment_tree[rt].isall = segment_tree[rt << 1].isall & segment_tree[rt << 1 | 1].isall;
}
void build(int rt, int l, int r)
{
segment_tree[rt].l = l;
segment_tree[rt].r = r;
if (l == r) {
segment_tree[rt].num =a[l];
if (segment_tree[rt].num == 1 || segment_tree[rt].num == 2) {
segment_tree[rt].isall = 1;
}
return ;
}
int mid = (l + r) >> 1;
build(rt << 1, l, mid);
build(rt << 1 | 1, mid + 1, r);
pushup(rt);
}
void update(int rt, int l, int r)
{
if (l <= segment_tree[rt].l && r >= segment_tree[rt].r && segment_tree[rt].isall) {
return;
}
if (segment_tree[rt].l == segment_tree[rt].r) {
segment_tree[rt].num = d[segment_tree[rt].num];
if (segment_tree[rt].num == 1 || segment_tree[rt].num == 2) {
segment_tree[rt].isall = 1;
}
return ;
} else {
int mid = (segment_tree[rt].l + segment_tree[rt].r) >> 1;
if (mid >= l) {
update(rt << 1, l, r);
}
if (mid < r) {
update(rt << 1 | 1, l, r);
}
pushup(rt);
}
}
ll query(int rt, int l, int r)
{
if (segment_tree[rt].l >= l && segment_tree[rt].r <= r) {
ll res = 0ll;
res += segment_tree[rt].num;
return res;
}
int mid = (segment_tree[rt].l + segment_tree[rt].r) >> 1;
ll res = 0ll;
if (mid >= l) {
res += query(rt << 1, l, r);
}
if (mid < r) {
res += query(rt << 1 | 1, l, r);
}
return res;
}
int main()
{
//freopen("D:\code\text\input.txt","r",stdin);
//freopen("D:\code\text\output.txt","w",stdout);
prepare();
int n, m;
du2(n, m);
repd(i, 1, n) {
du1(a[i]);
}
build(1, 1, n);
repd(i, 1, m) {
int op; int l, r;
du3(op, l, r);
if (op == 1) {
update(1, l, r);
} else {
printf("%lld
", query(1, l, r));
}
}
return 0;
}
inline void getInt(int *p)
{
char ch;
do {
ch = getchar();
} while (ch == ' ' || ch == '
');
if (ch == '-') {
*p = -(getchar() - '0');
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 - ch + '0';
}
} else {
*p = ch - '0';
while ((ch = getchar()) >= '0' && ch <= '9') {
*p = *p * 10 + ch - '0';
}
}
}