hdu 4777

题意：给定一个长度为n的序列，然后m次询问, 每个询问求[l, r]区间的与其他数都互质的数的个数

分析：非常感谢叉姐提供的思路，首先对于每一个a[i]我们可以求出 left[i],  right[i]表示这段区间的数与a[i]都互质，对于询问的区间我们按照左端点从小到大排序进行离线操作，对于到达left[i]时我们将[i, right[i]] +1, 离开i时将[i, right[i]] -1，这里可以用线段树进行维护

#include 

const int N = 200000 + 5;

int n, m;
int a[N], left[N], right[N];
int sum[N << 2], mark[N << 2];
int answer[N], pos[N];
std::vector  buffer[N];
std::vector  factors[N];
std::pair , int> p[N];

void prepare() {
    for (int i = 2; i < N; i ++) if ((int)factors[i].size() == 0) {
        for (int j = i; j < N; j += i) {
            factors[j].push_back(i);
        }
    }
}

void up(int o) {
    sum[o] = sum[o << 1] + sum[o << 1 | 1];
}

void down(int o, int l, int r) {
    int mid = (l + r) >> 1;
    sum[o << 1] += mark[o] * (mid - l + 1);
    sum[o << 1 | 1] += mark[o] * (r - mid);
    mark[o << 1] += mark[o];
    mark[o << 1 | 1] += mark[o];
    mark[o] = 0;
}

void update(int o, int l, int r, int L, int R, int x) {
    if (L <= l && R >= r) {
        mark[o] += x;
        sum[o] += x * (r - l + 1);
        return ;
    }
    if (mark[o] != 0) {
        down(o, l, r);
    }
    int mid = (l + r) >> 1;
    if (L <= mid) {
        update(o << 1, l, mid, L, R, x);
    }
    if (R > mid) {
        update(o << 1 | 1, mid + 1, r, L, R, x);
    }
    up(o);
}

int query(int o, int l, int r, int x) {
    if (l == r) {
        return sum[o];
    }
    if (mark[o] != 0) {
        down(o, l, r);
    }
    int mid = (l + r) >> 1;
    if (x <= mid) {
        return query(o << 1, l, mid, x);
    } else {
        return query(o << 1 | 1, mid + 1, r, x);
    }
}

void work() {
    for (int i = 1; i <= n; i ++) {
        buffer[i].clear();    
    }
    for (int i = 1; i <= n; i ++) {
        buffer[left[i]].push_back(i);
    }
    int now = 0;
    for (int i = 1; i <= 4 * n; i ++) {
        sum[i] = 0;
        mark[i] = 0;
    }
    for (int i = 1; i <= n; i ++) {
        for (int j = 0; j < (int)buffer[i].size(); j ++) {
            int l = buffer[i][j];
            int r = right[buffer[i][j]];
            update(1, 1, n, l, r, 1);
        }
        while (now < m && p[now].first.first == i) {
            answer[p[now].second] = query(1, 1, n, p[now].first.second);
            now ++;
        }
        if (now == m) {
            return ;
        }
        update(1, 1, n, i, right[i], -1);
    }
}

int main() {
    prepare();
    while (scanf("%d%d", &n, &m) == 2 && n + m) {
        for (int i = 1; i <= n; i ++) {
            scanf("%d", &a[i]);
        }
        int maxv = *std::max_element(a + 1, a + n + 1);
        std::fill(pos, pos + maxv + 1, 0);
        for (int i = 1; i <= n; i ++) {
            left[i] = 1;
            for (int j = 0; j < (int)factors[a[i]].size(); j ++) {
                int x = factors[a[i]][j];
                left[i] = std::max(left[i], pos[x] + 1);
                pos[x] = i;
            }
        }
        std::fill(pos, pos + maxv + 1, n + 1);
        for (int i = n; i >= 1; i --) {
            right[i] = n;
            for (int j = 0; j < (int)factors[a[i]].size(); j ++) {
                int x = factors[a[i]][j];
                right[i] = std::min(right[i], pos[x] - 1);
                pos[x] = i;
            }
        }
        for (int i = 0; i < m; i ++) {
            int l, r;
            scanf("%d%d", &l, &r);
            p[i].first = {l, r};
            p[i].second = i;
        }
        std::sort(p, p + m);
        work();
        for (int i = 0; i < m; i ++) {
            printf("%d\n", answer[i]);
        }
    }
    return 0;
}