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\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

CF1114F Please, another Queries on Array?

LuoGu: CF1114F Please, another Queries on Array?

CF: F. Please, another Queries on Array?

区间操作和查询考虑线段树。考察 $\varphi$ 函数的计算,我们不要知道一个数的质数因子的指数,只需要知道是否包含某个质数因子,注意到 $x$ 和 $a _ i$ 都很小,$300$ 以内只有 $62$ 个质数,考虑状压;另外维护区间乘积即可。

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#include <cstdio>
using namespace std;
template <class Type>
void read (Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 1) + (x << 3) + c - '0';
flag && (x = ~x + 1);
}
template <class Type, class ...rest>
void read (Type &x, rest &...y) { read(x), read(y...); }
template <class Type>
void write (Type x)
{
x < 0 && (putchar('-'), x = ~x + 1);
x > 9 && (write(x / 10), 0);
putchar('0' + x % 10);
}
typedef long long LL;
const int N = 4e5 + 10, M = 62, mod = 1e9 + 7, p[] = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293 };
int n, q, w[N], fp[N];
int binpow (int b, int k = mod - 2)
{
int res = 1;
for (; k; k >>= 1, b = (LL)b * b % mod)
if (k & 1) res = (LL)res * b % mod;
return res;
}
struct Data
{
LL s; int t;
Data (LL _s = 0, int _t = 1) { s = _s, t = _t; }
Data operator + (const Data &_) const { Data res = *this; return res += _; }
Data& operator += (const Data &_) { s |= _.s, t = (LL)t * _.t % mod; return *this; }
Data operator ^ (const int &k) const { Data res = *this; return res ^= k; }
Data& operator ^= (const int &k) { t = binpow(t, k); return *this; }
};
struct Node { Data s, t; int l, r; } tr[N << 2];
LL calc (int x)
{
LL res = 0;
for (int i = 0; i < M; ++i)
if (x % p[i] == 0) res |= 1ll << i;
return res;
}
void add (int x, Data t) { tr[x].s += t ^ tr[x].r - tr[x].l + 1, tr[x].t += t; }
void pushdown (int x)
{
add(x << 1, tr[x].t), add(x << 1 | 1, tr[x].t);
tr[x].t = Data();
}
void build (int l = 1, int r = n, int x = 1)
{
tr[x].l = l, tr[x].r = r;
if (l == r) return void(tr[x].s = Data(calc(w[l]), w[l]));
int mid = l + r >> 1;
build(l, mid, x << 1), build(mid + 1, r, x << 1 | 1);
tr[x].s = tr[x << 1].s + tr[x << 1 | 1].s;
}
void modify (int l, int r, Data k, int x = 1)
{
if (tr[x].l >= l && tr[x].r <= r) return add(x, k);
pushdown(x);
int mid = tr[x].l + tr[x].r >> 1;
if (l <= mid) modify(l, r, k, x << 1);
if (r > mid) modify(l, r, k, x << 1 | 1);
tr[x].s = tr[x << 1].s + tr[x << 1 | 1].s;
}
Data query (int l, int r, int x = 1)
{
if (tr[x].l >= l && tr[x].r <= r) return tr[x].s;
pushdown(x);
int mid = tr[x].l + tr[x].r >> 1;
if (r <= mid) return query(l, r, x << 1);
if (l > mid) return query(l, r, x << 1 | 1);
return query(l, r, x << 1) + query(l, r, x << 1 | 1);
}
int main ()
{
for (int i = 0; i < M; ++i)
fp[i] = (p[i] - 1ll) * binpow(p[i]) % mod;
read(n, q);
for (int i = 1; i <= n; ++i) read(w[i]);
build();
for (char op[10]; q; --q)
{
scanf("%s", op);
if (op[0] == 'M')
{
int l, r, x; read(l, r, x);
modify(l, r, Data(calc(x), x));
}
else if (op[0] == 'T')
{
int l, r; read(l, r);
Data k = query(l, r);
for (int i = 0; i < M; ++i)
if (k.s >> i & 1) k.t = (LL)k.t * fp[i] % mod;
write((k.
t + mod) % mod), puts("");
}
}
return 0;
}