query

Given a permutation pp of length nn, you are asked to answer mm queries, each query can be represented as a pair (l ,r )(l,r), you need to find the number of pair(i ,j)(i,j) such that l \le i < j \le rl≤i<j≤r and \min(p_i,p_j) = \gcd(p_i,p_j )min(pi​,pj​)=gcd(pi​,pj​).

Input

There is two integers n(1 \le n \le 10^5)n(1≤n≤105), m(1 \le m \le 10^5)m(1≤m≤105) in the first line, denoting the length of pp and the number of queries.

In the second line, there is a permutation of length nn, denoting the given permutation pp. It is guaranteed that pp is a permutation of length nn.

For the next mm lines, each line contains two integer l_ili​ and r_i(1 \le l_i \le r_i \le n)ri​(1≤li​≤ri​≤n), denoting each query.

Output

For each query, print a single line containing only one integer which denotes the number of pair(i,j)(i,j).

样例输入复制

3 2
1 2 3
1 3
2 3

样例输出复制

2
0

#include 

#define lowbit(x) x&(-x)
using namespace std;
typedef long long ll;
const int maxn = 3e6 + 10;
const ll mod = 1e9 + 7;

struct BIT {
    int n, c[maxn];

    inline void init(int _n_) {
        memset(c, 0, sizeof(c));
        n = _n_ + 10;
    }

    inline void update(int pos, int val) {
        while (pos <= n) {
            c[pos] += val;
            pos += lowbit(pos);
        }
    }

    inline int query(int pos) {
        int res = 0;
        while (pos) {
            res += c[pos];
            pos -= lowbit(pos);
        }
        return res;
    }
} bit;

int res[maxn];

struct event {

    int op, l, r, id;

    bool operator<(const event &cur) const {
        if (l != cur.l)return l < cur.l;
        return op < cur.op;
    }

    inline void precess() {
        if (op == 0) {
            bit.update(r, 1);
        }
        else {
            int w = bit.query(r);
            if (op == 1) {
                res[id] += w;
            }
            else {
                res[id] -= w;
            }
        }
    }
};

int n, m;
int o[maxn], pos[maxn];
vector<event> e;

int main() {
#ifndef ONLINE_JUDGE
    freopen("1.txt", "r", stdin);
#endif
    scanf("%d%d", &n, &m);
    for (register int i = 1; i <= n; ++i) {
        scanf("%d", &o[i]);
        pos[o[i]] = i;
    }
    bit.init(n);
    for (register int i = 1; i <= n; ++i) {
        for (register int j = i + i; j <= n; j += i) {
            e.emplace_back(event{0, pos[i], pos[j], 0});
        }
    }
    for (register int i = 1, l, r; i <= m; ++i) {
        scanf("%d%d", &l, &r);
        --l;
        e.emplace_back(event{1, l, l, i});
        e.emplace_back(event{1, r, r, i});
        e.emplace_back(event{2, l, r, i});
        e.emplace_back(event{2, r, l, i});
    }
    sort(e.begin(), e.end());
    for (auto &cur:e) {
        cur.precess();
    }
    for (register int i = 1; i <= m; ++i) {
        printf("%d\n", res[i]);
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/czy-power/p/11483056.html