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 = ;

int st[MAXN][];
int n, t;

map<int, int> pos;
vector<int> f[MAXN];

inline int  gcd (int x, int y) {
    return y ==  ? x : gcd (y, x % y);
}

inline void ST() {
    for (int i = n - ; i > ; --i)
        for (int j = ; i + ( << j) <= n; j++)
            st[i][j] = gcd (gcd (st[i][j], st[i][j - ]), st[i + ( << j - )][j - ]);
}
inline int getgcd (int l, int r) {
    int tem = st[l][];
    for (int k = ; l + ( << k) <= r; k++)
        tem = gcd (gcd (tem, st[l][k]), st[r - ( << k) + ][k]);
    return tem;
}
int main() {
    ios::sync_with_stdio ();
    cin >> n;
    int tol = ;
    for (int i = ; i <= n; i++)  {
        cin >> st[i][];
        if (pos.find (st[i][]) == pos.end() ) tol++, pos[st[i][]] = tol;
        f[pos[st[i][]]].push_back (i);
    }
    ST();
    cin >> t;
    for (int i = , 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 +<< endl;
    }
}