Codeforces1114F Please, another Queries on Array?

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 = ; T f = ; char ch = getchar();
    while (ch < '' || ch > '') { if (ch == '-') f = -; ch = getchar(); }
    while (ch >= '' && ch <= '') { x = x *  + ch - ; ch = getchar() ;}
    x *= f;
}

const ll MOD = 1e9 + ;
const int N = 4e5 + ;
int prime[], prin, a[N], n;
ll inv[];
bool vis[];

ll qp(ll a, ll b) {
    ll res = ;
    while (b) {
        if (b & ) res = res * a % MOD;
        a = a * a % MOD;
        b >>= ;
    }
    return res;
}

inline ll getbin(ll x) {
    ll res = ;
    for (int i = ; i < prin; i++)
        if (x % prime[i] == ) res |= (1LL << i);
    return res;
}

struct SEG {
    ll appear[N << ], num[N << ], lazy[N << ], lazy_bin[N << ];
    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] = ;
        lazy_bin[p] = ;
    }
    void build(int p, int l, int r) {
        appear[p] = num[p] = lazy_bin[p] = ;
        lazy[p] = ;
        if (l == r) {
            num[p] = a[l];
            appear[p] = getbin(a[l]);
            return;
        }
        int mid = l + r >> ;
        build(lp, l, mid);
        build(rp, mid + , 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 + ) % MOD;
            return;
        }
        int mid = l + r >> ;
        pushdown(p, mid - l + , r - mid);
        if (x <= mid) update(lp, l, mid, x, y, v, bin);
        if (y > mid) update(rp, mid + , 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 >> ;
        pushdown(p, mid - l + , r - mid);
        if (x <= mid) query(lp, l, mid, x, y, v, bin);
        if (y > mid) query(rp, mid + , r, x, y, v, bin);
    }
} seg;

char s[];

int main() {
    //freopen("in.txt", "r", stdin);
    for (int i = ; i < ; i++) {
        if (!vis[i]) prime[prin++] = i;
        for (int j = ; j < prin && i * prime[j] < ; j++) {
            vis[i * prime[j]] = ;
            if (i % prime[j] == ) break;
        }
    }
    inv[] = ;
    for (int i = ; i < ; i++) inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD;
    read(n);
    int q; read(q);
    for (int i = ; i <= n; i++) read(a[i]);
    seg.build(, , n);
    while (q--) {
        scanf("%s", s);
        if (s[] == 'T') {
            int l, r;
            read(l); read(r);
            ll v = , bin = ;
            seg.query(, , n, l, r, v, bin);
            for (int i = ; i < prin; i++) {
                if (bin >> i & ) v = v * (prime[i] - ) % MOD * inv[prime[i]] % MOD;
            }
            printf("%lld\n", v);
        } else {
            int l, r, w;
            read(l); read(r); read(w);
            seg.update(, , n, l, r, w, getbin(w));
        }
    }
    return ;
}