标签：node 超过 template build inpu make lines text lld
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
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。
$ 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。
#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\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';
}
}
}
?Educational Codeforces Round 37-F.SUM and REPLACE （线段树，线性筛，收敛函数）
标签：node 超过 template build inpu make lines text lld
原文地址：https://www.cnblogs.com/qieqiemin/p/11617207.html
