Codeforces Round #538 (Div. 2) F. Please, another Queries on Array?

原题链接 F. Please, another Queries on Array?

这道题让求$\phi(\prod\limits_{i = l}^r a_i)$,然后我们化简一下。
设$P$是$\prod\limits_{i = l}^r a_i$质因子的集合，那么化简得：

$$\phi(\prod\limits_{i = l}^r a_i) = \prod\limits_{i = l}^r a_i \times \prod\limits_{p \in P} \cfrac{p-1}{p}$$

因此我们分成两部分计算，前边部分$\prod\limits_{i = l}^r a_i$很容易维护，后边部分
$\prod\limits_{p \in P} \cfrac{p-1}{p}$，我们知道一个数的欧拉函数只与有几个质因子有关，而$300$以内的质数只有$62$个，所有我们每次修改的时候只需要暴力枚举$62$个质数，然后状压存储即可。

// Problem: F. Please, another Queries on Array?
// Contest: Codeforces - Codeforces Round #538 (Div. 2)
// URL: https://codeforces.com/contest/1114/problem/F
// Memory Limit: 256 MB
// Time Limit: 5500 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

const int N = 4E5 + 10;
const LL Mod = 1e9 + 7;
int a[N];
int primes[310], cnt = 0, st[310];
char op[10];

void get_primes(int n) {
	for (int i = 2; i <= n; i++) {
		if (!st[i]) primes[cnt++] = i;
		for (int j = 0; primes[j] <= n / i; j++) {
			st[primes[j] * i] = 1; 
			if (i % primes[j] == 0) break;
		}
	}
}

struct SegMentTree {
	int l, r;
	LL products, mul, OR, tag;
}tr[N * 4];

LL qmi(LL a, LL b, LL p) {
	LL res = 1ll;
	while (b) {
		if (b & 1) res = res * a % p;
		b >>= 1;
		a = a * a % p;
	}
	return res;
}

void pushup(int u) {
	tr[u].products = tr[u << 1].products * tr[u << 1 | 1].products % Mod;
	tr[u].OR = tr[u << 1].OR | tr[u << 1 | 1].OR;
}

void pushdown(int u) {
	if (tr[u].mul == 1 || !tr[u].tag) return;
	auto &root = tr[u], &left = tr[u << 1], &right = tr[u << 1 | 1];
	left.mul = left.mul * root.mul % Mod; right.mul = right.mul * root.mul % Mod;
	left.products = left.products * qmi(root.mul, left.r - left.l + 1, Mod) % Mod;
	right.products = right.products * qmi(root.mul, right.r - right.l + 1, Mod) % Mod;
	left.tag = left.tag | root.tag; right.tag = right.tag | root.tag;
	left.OR = left.OR | root.tag; right.OR = right.OR | root.tag;
	root.mul = 1; root.tag = 0;	
}

void build(int u, int l, int r) {
	if (l == r) {
		tr[u] = {l, r, 1ll * a[l], 1ll};
		for (int i = 0; i < cnt; i++) {
			if (a[l] % primes[i] == 0) {
				tr[u].OR |= (1ll << i);
			}
		}
	} else {
		tr[u].l = l, tr[u].r = r, tr[u].mul = 1ll;
		//tr[u] = {l, r};
		int mid = l + r >> 1;
		build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);	
		pushup(u);
	}
}

void modify(int u, int l, int r, LL p, int x) {
	if (l <= tr[u].l && tr[u].r <= r) {
		tr[u].mul = tr[u].mul * 1ll * x % Mod;
		tr[u].products = tr[u].products * qmi(1ll * x, 1ll * (tr[u].r - tr[u].l + 1), Mod) % Mod;
		tr[u].OR |= p, tr[u].tag |= p;
		return;
	}
	pushdown(u);
	int mid = tr[u].l + tr[u].r >> 1;
	if (l <= mid) modify(u << 1, l, r, p, x);
	if (r > mid) modify(u << 1 | 1, l, r, p, x);
	pushup(u);
}

LL ask(int u, int l, int r) {
	if (l <= tr[u].l && tr[u].r <= r) {
		return tr[u].products;
	}
	pushdown(u);
	LL res = 1ll;
	int mid = tr[u].l + tr[u].r >> 1;
	if (l <= mid) res = res * ask(u << 1, l, r) % Mod;
	if (r > mid) res = res * ask(u << 1 | 1, l, r) % Mod;
	return res;
}

int ask_flag(int u, int l, int r, int i) {
	if (l <= tr[u].l && tr[u].r <= r) {
		return (tr[u].OR >> i & 1);
	}
	pushdown(u);
	LL res = 0ll;
	int mid = tr[u].l + tr[u].r >> 1;
	if (l <= mid) res += ask_flag(u << 1, l, r, i);
	if (r > mid) res += ask_flag(u << 1 | 1, l, r, i);
	return res;
}

int main() {
	get_primes(300);
	//cout << cnt << endl;
	int n, q;
	scanf("%d%d", &n, &q);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
	}
	
	build(1, 1, n);
	
	while (q--) {
		int l, r, x;
		//cin >> s;
		scanf("%s%d%d", &op, &l, &r);
		if (op[0] == 'M') {
			scanf("%d", &x);
			LL p = 0;
			for (int i = 0; i < cnt; i++) {
				if (x % primes[i] == 0) {
					p |= (1ll << i);
				}
			}
			modify(1, l, r, p, x);
		} else if (op[0] == 'T') { 
			LL res = ask(1, l, r);
			for (int i = 0; i < cnt; i++) {
				int p = primes[i];
				if (ask_flag(1, l, r, i)) {
					res = res * (p - 1) % Mod;
					res = res * qmi(p, Mod - 2, Mod) % Mod;
				}
			}
			printf("%lld\n", res);
		}
	}
	
    return 0;
}