题目链接：http://codeforces.com/problemset/problem/1114/F
题意：序列$a$，有两种操作，1 区间里的数同时乘上$x$ 2 求区间的积的欧拉函数
线段树好题。
思路：最直观的思路是，线段树每个节点维护的是一个数组，表示这个数每个素因子及出现的次数，欧拉函数值用矩阵快速幂求解即可。但是这样空间不大够，而且复杂度多个常数，因此不大行。
发现300以内的素数刚好有62个，那么可以用一个long long的二进制数来表示这个数有那些素因子出现，在维护一下区间积，欧拉函数值可以用公式
$\varphi \left( n\right) =n\cdot \dfrac {\prod \left( p_{i}-1\right) }{\prod p_{i}}$
lazy_tag有两个，一个是异或，一个是乘积。我刚开始没用异或，准备直接用乘积来得到异或值，但是因为会取模，所以乘积标记不能用来得到位上的值。
#include <bits/stdc++.h> #define lp p << 1 #define rp p << 1 | 1 #define ll long long using namespace std; template<typename T> inline void read(T &x) { x = 0; T f = 1; char ch = getchar(); while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); } while (ch >= '0' && ch <= '9') { x = x * 10 + ch - 48; ch = getchar() ;} x *= f; } const ll MOD = 1e9 + 7; const int N = 4e5 + 10; int prime[70], prin, a[N], n; ll inv[300]; bool vis[300]; ll qp(ll a, ll b) { ll res = 1; while (b) { if (b & 1) res = res * a % MOD; a = a * a % MOD; b >>= 1; } return res; } inline ll getbin(ll x) { ll res = 0; for (int i = 0; i < prin; i++) if (x % prime[i] == 0) res |= (1LL << i); return res; } struct SEG { ll appear[N << 4], num[N << 4], lazy[N << 4], lazy_bin[N << 4]; void pushup(int p) { appear[p] |= (appear[lp] | appear[rp]); num[p] = num[lp] * num[rp] % MOD; } void pushdown(int p, int llen, int rlen) { lazy[lp] = lazy[lp] * lazy[p] % MOD; lazy[rp] = lazy[rp] * lazy[p] % MOD; lazy_bin[lp] |= lazy_bin[p]; lazy_bin[rp] |= lazy_bin[p]; appear[lp] |= lazy_bin[lp]; appear[rp] |= lazy_bin[rp]; num[lp] = num[lp] * qp(lazy[p], llen) % MOD; num[rp] = num[rp] * qp(lazy[p], rlen) % MOD; lazy[p] = 1; lazy_bin[p] = 0; } void build(int p, int l, int r) { appear[p] = num[p] = lazy_bin[p] = 0; lazy[p] = 1; if (l == r) { num[p] = a[l]; appear[p] = getbin(a[l]); return; } int mid = l + r >> 1; build(lp, l, mid); build(rp, mid + 1, r); pushup(p); } void update(int p, int l, int r, int x, int y, ll v, ll bin) { if (x <= l && r <= y) { lazy_bin[p] |= bin; appear[p] |= bin; lazy[p] = lazy[p] * v % MOD; num[p] = num[p] * qp(v, r - l + 1) % MOD; return; } int mid = l + r >> 1; pushdown(p, mid - l + 1, r - mid); if (x <= mid) update(lp, l, mid, x, y, v, bin); if (y > mid) update(rp, mid + 1, r, x, y, v, bin); pushup(p); } void query(int p, int l, int r, int x, int y, ll &v, ll &bin) { if (x <= l && r <= y) { v = v * num[p] % MOD; bin |= appear[p]; return; } int mid = l + r >> 1; pushdown(p, mid - l + 1, r - mid); if (x <= mid) query(lp, l, mid, x, y, v, bin); if (y > mid) query(rp, mid + 1, r, x, y, v, bin); } } seg; char s[20]; int main() { //freopen("in.txt", "r", stdin); for (int i = 2; i < 300; i++) { if (!vis[i]) prime[prin++] = i; for (int j = 0; j < prin && i * prime[j] < 300; j++) { vis[i * prime[j]] = 1; if (i % prime[j] == 0) break; } } inv[1] = 1; for (int i = 2; i < 300; i++) inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD; read(n); int q; read(q); for (int i = 1; i <= n; i++) read(a[i]); seg.build(1, 1, n); while (q--) { scanf("%s", s); if (s[0] == 'T') { int l, r; read(l); read(r); ll v = 1, bin = 0; seg.query(1, 1, n, l, r, v, bin); for (int i = 0; i < prin; i++) { if (bin >> i & 1) v = v * (prime[i] - 1) % MOD * inv[prime[i]] % MOD; } printf("%lld\n", v); } else { int l, r, w; read(l); read(r); read(w); seg.update(1, 1, n, l, r, w, getbin(w)); } } return 0; }