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

題意：

給你一個含有n個數的數組，和m個操作

操作1：將l~r中每一個數\(a[i]\)變成 \(d(a[i])\)

? 其中$ d(x)$ 是約數個數函數。

操作2： 求l~r的a[i] 的sum和。

思路：

$ d(x)$ 約數個數函數可以利用線性篩預處理處理。

又因為 \(d(2)=2\) 和 \(d(1)=1\) 操作1對a[i]等于1或者2沒有影響。

那么我們可以對一個區間中全都是1或者2不更新操作。

同時 \(d(x)\) 是收斂函數， 在1e6 的范圍內，最多不超過5次改變就會收斂到1或2.

所以更新操作可以暴力解決，

同時用線段樹維護即可。

代碼：

#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), '\0', 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("\n");}}} 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("\n");}}}inline void getInt(int *p); const int maxn = 1000010; const int inf = 0x3f3f3f3f; /*** TEMPLATE CODE * * STARTS HERE ***/ // d(n)表示n的約數個數和 // prime[i]表示第i個質數 //num[i]表示i的最小質因子出現次數 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\n",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\n", query(1, l, r));}}return 0; }inline void getInt(int *p) {char ch;do {ch = getchar();} while (ch == ' ' || ch == '\n');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';}} }

轉載于:https://www.cnblogs.com/qieqiemin/p/11617207.html

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