hdu4777-Rabbit Kingdom

题意：求区间内与其他任何数都互质的数的个数。

题解：求出每个数左右互质的边界。然后对询问排序，通过树状数组求解。

讲道理真的好难啊= =

http://blog.csdn.net/dyx404514/article/details/15507209 这个博客讲的最清楚（竟然是戴神的博客=。= 听过戴神讲splay，现在还不会……

#include <bits/stdc++.h>

using namespace std;

const int N = 200005;

/***素数打表***/
int p[N];
int is_p[N+1];
int cnt_p;
void get_p() {
    for (int i = 0; i <= N; ++i) is_p[i] = 1;
    cnt_p = is_p[0] = is_p[1] = 0;
    for (int i = 2; i <= N; ++i) {
        if (is_p[i]) {
            p[cnt_p++] = i;
            for (int j = i * 2; j <= N; j += i) is_p[j] = 0;
        }
    }
}
/***因数分解***/
int fac[N];     // 记录出现的因数
int cal_fac(int x) {
    int tmp = x;
    int idx = 0;
    int i;
    for (i = 0; p[i] <= tmp/p[i]; ++i) {
        if (tmp%p[i]) continue;
        while (tmp%p[i] == 0) tmp /= p[i];
        fac[idx++] = p[i];
    }
    if (tmp != 1) fac[idx++] = tmp;
    return idx;
}
/***计算每个数互质的最左最右区间***/
int l[N], r[N];
int adj[N];
int a[N];
void cal_inv(int n) {
    for (int i = 0; i < N; ++i) adj[i] = 0;
    for (int i = 1; i <= n; ++i) {
        int cnt = cal_fac(a[i]);
        l[i] = 0;
        for (int j = 0; j < cnt; ++j) {
            l[i] = max(l[i], adj[fac[j]]);
            adj[fac[j]] = i;
        }
    }
    for (int i = 0; i < N; ++i) adj[i] = n+1;
    for (int i = n; i >= 1; --i) {
        r[i] = n+1;
        int cnt = cal_fac(a[i]);
        for (int j = 0; j < cnt; ++j) {
            r[i] = min(r[i], adj[fac[j]]);
            adj[fac[j]] = i;
        }
    }
}

struct node {
    int l, r, id;
    bool operator<(const node x) const { return l < x.l; }
} qry[N];

int bit[N];
int lowbit(int x) { return x&-x; }
void add(int x, int v, int n) { while(x<=n) bit[x]+=v,x+=lowbit(x); }
int sum(int x) { int ans=0; while(x) ans+=bit[x],x-=lowbit(x); return ans; }

vector<int> lb[N]; // lb[i] 以i为左边界的数
int ans[N];
int main() {
//freopen("in.txt", "r", stdin);
    get_p();
    int n, k;
    while (~scanf("%d%d", &n, &k)) {
        if (n == 0) break;
        for (int i = 1; i <= n; ++i) {
            scanf("%d", a+i);
        }
        cal_inv(n);

        for (int i = 0; i < k; ++i) {
            scanf("%d%d", &qry[i].l, &qry[i].r);
            qry[i].id = i;
        }
        sort(qry, qry+k);
        for (int i = 0; i <= n; ++i) lb[i].clear();
        memset(bit, 0, sizeof bit);
        for (int i = 1; i <= n; ++i) {
            if (!l[i]) { add(i, 1, n); add(r[i], -1, n); }
            else lb[l[i]].push_back(i);
        }
        int pos = 1;
        for (int i = 0; i < k; ++i) {
            while (qry[i].l > pos) {
                add(pos, -1, n);
                add(r[pos], 1, n);
                for (unsigned j = 0; j < lb[pos].size(); ++j) {
                    int x = lb[pos][j];
                    add(x, 1, n);
                    add(r[x], -1, n);
                }
                pos++;
            }
            ans[qry[i].id] = sum(qry[i].r)-sum(qry[i].l-1);
        }
        for (int i = 0; i < k; ++i) {
            printf("%d\n", ans[i]);
        }
    }

    return 0;
}