Codeforces 594D REQ 线段树

REQ

把询问离线, 我们从n 到 1遍历过去的时候, 把(1 - 1 / p)乘在最靠近当前位置的地方, 然后区间求乘积就好啦。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ull unsigned long long
using namespace std;

const int N = 1e6 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;

struct SegmentTree {
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
    int a[N << 2];
    void build(int *b, int l, int r, int rt) {
        if(l == r) {
            a[rt] = b[l];
            return;
        }
        int mid = l + r >> 1;
        build(b, lson); build(b, rson);
        a[rt] = 1ll * a[rt << 1] * a[rt << 1 | 1] % mod;
    }
    void update(int p, int val, int l, int r, int rt) {
        if(l == r) {
            a[rt] = 1ll * a[rt] * val % mod;
            return;
        }
        int mid = l + r >> 1;
        if(p <= mid) update(p, val, lson);
        else update(p, val, rson);
        a[rt] = 1ll * a[rt << 1] * a[rt << 1 | 1] % mod;
    }
    int query(int L, int R, int l, int r, int rt) {
        if(l >= L && r <= R) return a[rt];
        int mid = l + r >> 1;
        if(R <= mid) return query(L, R, lson);
        else if(L > mid) return query(L, R, rson);
        else return (1ll * query(L, R, lson) * query(L, R, rson)) % mod;
    }
};

int n, q, a[N], mul[N], del[N], ans[N], Map[N];
vector<int> prime;
vector<int> fac[N];
vector<PII> qus[N];
SegmentTree seg;
int Power(int a, int b) {
    int ans = 1;
    while(b) {
        if(b & 1) ans = 1ll * ans * a % mod;
        a = 1ll * a * a % mod; b >>= 1;
    }
    return ans;
}

int main() {
    for(int i = 2; i < N; i++) {
        if(SZ(fac[i])) continue;
        prime.push_back(i);
        for(int j = i; j < N; j += i)
            fac[j].push_back(i);
    }
    for(auto& x : prime) {
        mul[x] = (1 - Power(x, mod - 2) + mod) % mod;
        del[x] = Power(mul[x], mod - 2);
    }
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
    seg.build(a, 1, n, 1);
    scanf("%d", &q);
    for(int i = 1; i <= q; i++) {
        int L, R; scanf("%d%d", &L, &R);
        qus[L].push_back(mk(R, i));
    }
    for(int i = n; i >= 1; i--) {
        for(auto& t : fac[a[i]]) {
            if(Map[t]) seg.update(Map[t], del[t], 1, n, 1);
            seg.update(i, mul[t], 1, n, 1);
            Map[t] = i;
        }
        for(auto& t : qus[i]) {
            ans[t.se] = seg.query(i, t.fi, 1, n, 1);
        }
    }
    for(int i = 1; i <= q; i++) printf("%d\n", ans[i]);
    return 0;
}

/*
*/

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转载自www.cnblogs.com/CJLHY/p/10499001.html
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