Codeforces 474F - Ant colony

注意到每个区间生存下来的蚂蚁的长度等于区间的gcd

于是可以先预处理出区间的gcd

然后二分查找就好了

预处理gcd我这里用的是倍增法

总的时间复杂度O（NlogN）

/*

       Cf 271F

       倍增求区间GCD

       对下标二分

       时间复杂度O(NlogN)

*/

#include <iostream>

#include <algorithm>

#include <vector>

#include <map>

using namespace std;



const int MAXN = 100009;



int st[MAXN][20];

int n, t;



map<int, int> pos;

vector<int> f[MAXN];



inline int  gcd (int x, int y) {

    return y == 0 ? x : gcd (y, x % y);

}



inline void ST() {

    for (int i = n - 1; i > 0; --i)

        for (int j = 1; i + (1 << j) <= n; j++)

            st[i][j] = gcd (gcd (st[i][j], st[i][j - 1]), st[i + (1 << j - 1)][j - 1]);

}

inline int getgcd (int l, int r) {

    int tem = st[l][0];

    for (int k = 0; l + (1 << k) <= r; k++)

        tem = gcd (gcd (tem, st[l][k]), st[r - (1 << k) + 1][k]);

    return tem;

}

int main() {

    ios::sync_with_stdio (0);

    cin >> n;

    int tol = 0;

    for (int i = 1; i <= n; i++)  {

        cin >> st[i][0];

        if (pos.find (st[i][0]) == pos.end() ) tol++, pos[st[i][0]] = tol;

        f[pos[st[i][0]]].push_back (i);

    }

    ST();

    cin >> t;

    for (int i = 1, l, r; i <= t; i++) {

        cin >> l >> r;

        int key = getgcd (l, r), k = pos[key];

        int d = upper_bound (f[k].begin(), f[k].end(), r) - lower_bound (f[k].begin(), f[k].end(), l);

        cout << r - l  - d +1<< endl;

    }

}

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