HDU 5634 Rikka with Phi

Problem Description

Rikka and Yuta are interested in Phi function (which is known as Euler's totient function).

Yuta gives Rikka an array  A[1..n]  of positive integers, then Yuta makes  m  queries. 

There are three types of queries: 

1lr  

Change  A[i]  into  φ(A[i]) , for all  i∈[l,r] .

2lrx  

Change  A[i]  into  x , for all  i∈[l,r] .

3lr  

Sum up  A[i] , for all  i∈[l,r] .

Help Rikka by computing the results of queries of type 3.

 

Input

The first line contains a number  T(T≤100)  ——The number of the testcases. And there are no more than 2 testcases with  n>105

For each testcase, the first line contains two numbers  n,m(n≤3×105,m≤3×105) 。

The second line contains  n  numbers  A[i]

Each of the next  m  lines contains the description of the query. 

It is guaranteed that  1≤A[i]≤107  At any moment.

 

Output

For each query of type 3, print one number which represents the answer.

 

Sample Input

   
   
   
   

    
    
    
    1
10 10
56 90 33 70 91 69 41 22 77 45
1 3 9
1 1 10
3 3 8
2 5 6 74
1 1 8
3 1 9
1 2 10
1 4 9
2 8 8 69
3 3 9
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    80
122
86
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    因为欧拉函数下降是logn的，所以直接线段树暴力更新即可
   
   
   
   

#include<cstdio>
#include<cstring>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<functional>
using namespace std;
typedef unsigned long long ull;
typedef long long LL;
const int maxn = 1e7 + 10;
int T, euler[maxn], n, m, x, l, r, c;

void Init(){
	euler[1] = 1;
	for (int i = 2; i<maxn; i++) euler[i] = i;
	for (int i = 2; i<maxn; i++)
		if (euler[i] == i)
			for (int j = i; j<maxn; j += i) euler[j] = euler[j] / i*(i - 1);
}

struct Tree
{
	const static int maxn = 1e6 + 2e5 + 10;
	int F[maxn];
	LL sum[maxn];
	void build(int x, int l, int r)
	{
		if (l == r) { scanf("%d", &F[x]); sum[x] = F[x]; return; }
		int mid = l + r >> 1;
		build(x << 1, l, mid);
		build(x << 1 | 1, mid + 1, r);
		sum[x] = sum[x << 1] + sum[x << 1 | 1];
		if (F[x << 1] == F[x << 1 | 1]) F[x] = F[x << 1]; else F[x] = 0;
	}
	void pushdown(int x, int l, int mid, int r)
	{
		F[x << 1] = F[x << 1 | 1] = F[x];
		sum[x << 1] = (LL)F[x] * (mid - l + 1);
		sum[x << 1 | 1] = (LL)F[x] * (r - mid);
		F[x] = 0;
	}
	void make(int x, int l, int r, int ll, int rr, int c)
	{
		if (ll <= l&&r <= rr) { F[x] = c; sum[x] = (LL)(r - l + 1)*c; return; }
		int mid = l + r >> 1;
		if (F[x]) pushdown(x, l, mid, r);
		if (ll <= mid) make(x << 1, l, mid, ll, rr, c);
		if (rr > mid) make(x << 1 | 1, mid + 1, r, ll, rr, c);
		sum[x] = sum[x << 1] + sum[x << 1 | 1];
		if (F[x << 1] == F[x << 1 | 1]) F[x] = F[x << 1]; else F[x] = 0;
	}
	void get(int x, int l, int r, int ll, int rr)
	{
		if (ll <= l&&r <= rr)
		{
			if (F[x]) { F[x] = euler[F[x]]; sum[x] = (LL)F[x] * (r - l + 1); return; }
			int mid = l + r >> 1;
			get(x << 1, l, mid, ll, rr);
			get(x << 1 | 1, mid + 1, r, ll, rr);
			sum[x] = sum[x << 1] + sum[x << 1 | 1];
			if (F[x << 1] == F[x << 1 | 1]) F[x] = F[x << 1]; else F[x] = 0;
			return;
		}
		int mid = l + r >> 1;
		if (F[x]) pushdown(x, l, mid, r);
		if (ll <= mid) get(x << 1, l, mid, ll, rr);
		if (rr > mid) get(x << 1 | 1, mid + 1, r, ll, rr);
		sum[x] = sum[x << 1] + sum[x << 1 | 1];
		if (F[x << 1] == F[x << 1 | 1]) F[x] = F[x << 1]; else F[x] = 0;
	}
	LL putout(int x, int l, int r, int ll, int rr)
	{
		if (ll <= l&&r <= rr) return sum[x];
		int mid = l + r >> 1;
		LL ans = 0;
		if (F[x]) pushdown(x, l, mid, r);
		if (ll <= mid) ans+=putout(x << 1, l, mid, ll, rr);
		if (rr > mid) ans+=putout(x << 1 | 1, mid + 1, r, ll, rr);
		return ans;
	}
}tree;

int main()
{
	Init();
	cin >> T;
	while (T--)
	{
		scanf("%d%d", &n, &m);
		tree.build(1, 1, n);
		while (m--)
		{
			scanf("%d%d%d", &x, &l, &r);
			if (x == 1) tree.get(1, 1, n, l, r);
			if (x == 2) scanf("%d", &c), tree.make(1, 1, n, l, r, c);
			if (x == 3) printf("%lld\n", tree.putout(1, 1, n, l, r));
		}
	}
	return 0;
}