• 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
    ", ans[i]);
        return 0;
    }
    
    /*
    */
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  • 原文地址:https://www.cnblogs.com/CJLHY/p/10499001.html
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