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.
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.
For each SUM query print the answer to it.
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
30
13
4
22
题目分析 ：题目说了一种操作，就是将一个数变成它的因数的个数，第二种操作是查询一段区间的和，首先你观察它的查询和更改的操作次数有 3e5 次，查询的话肯定是没有问题的， logn 的复杂度，但是你更改呢？每个数都改，一定会超时，那么要看下这些被改的数有什么特征，一个很大的数，经过一次质因数操作，会变成一个相对很小的数，重复此过程，会变得越来越小，直到变成2，那么我线段树可以优化的地方是不就在这里，设一个标记，就可以了
代码示例 ：
const ll maxn = 1e6+5;
const ll maxn2 = 3e5+5;
const double pi = acos(-1.0);
const ll inf = 0x3f3f3f3f;
#define lson k<<1
#define rson k<<1|1
ll cnt[maxn];
struct node
{
ll l, r;
ll sum;
ll pt;
}t[maxn2<<2];
void init() {
for(int i = 1; i <= 1000000; i++) cnt[i] = 2;
cnt[1] = 1;
for(int i = 2; i <= 1000000; i++){
for(int j = i+i; j <= 1000000; j += i) cnt[j]++;
}
}
void pushup(int k){
t[k].sum = t[lson].sum + t[rson].sum;
if (t[lson].pt && t[rson].pt) t[k].pt = 1;
}
void build(ll l, ll r, ll k){
t[k].l = l; t[k].r = r;
t[k].pt = 0;
if (l == r) {
scanf("%lld", &t[k].sum);
if (t[k].sum == cnt[t[k].sum]) t[k].pt = 1;
return;
}
ll m = (l + r) >> 1;
build(l, m, lson);
build(m+1, r, rson);
pushup(k);
}
ll query(ll l, ll r, ll k){
ll s = 0;
if (l <= t[k].l && t[k].r <= r){
s += t[k].sum;
return s;
}
ll m = (t[k].l + t[k].r) >> 1;
if (l <= m) s += query(l, r, lson);
if (r > m) s += query(l, r, rson);
return s;
}
void update(ll l, ll r, ll k){
if (t[k].pt) return;
if (t[k].l == t[k].r) {
t[k].sum = cnt[t[k].sum];
if (t[k].sum == cnt[t[k].sum]) t[k].pt = 1;
//printf("***** %lld\n", cnt[x]);
return;
}
ll m = (t[k].l + t[k].r) >> 1;
if (l <= m) update(l, r, lson);
if (r > m) update(l, r, rson);
pushup(k);
}
int main() {
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
ll n, m;
ll pt, l, r;
init();
//for(ll i = 1; i <= 100; i++) printf("%lld %lld\n", i, cnt[i]);
cin >> n >> m;
build(1, n, 1);
//printf("%lld\n", query(1, 7, 1));
for(ll i = 1; i <= m; i++){
scanf("%lld%lld%lld", &pt, &l, &r);
if (pt == 1) {
update(l, r, 1);
}
else {
printf("%lld\n", query(l, r, 1));
}
}
return 0;
}
