hdu 4777 Rabbit Kingdom(树状数组)

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题目链接：hdu 4777 Rabbit Kingdom

题目大意：一个兔子王国，有N只兔子，每只兔子有一个重量，如果两只兔子的重量不互质，那么就会干架，现在国王想将l r之间的兔子关进监狱，它想知道会有多少只兔子不会和别的兔子干架。

解题思路：预处理出每只兔子的L，R表示向左和向右最近会与该兔子发生冲突的兔子，预处理的时候只要将每只兔子的重量分解成质因子后遍历两遍。
对于询问，将询问按照右区间排序，碰到i，则L位置+1，碰到R，则i位置+1，L位置-1。（如果L ≤ l && r ≤ R，那么兔子在这个l，r询问中是不会与其他兔子冲突）
这样l~r区间统计的即为会打架的兔子。

#include <cstdio>#include <cstring>#include <cmath>#include <vector>#include <algorithm>using namespace std;const int maxn = 200005;int np, pri[maxn], vis[maxn];void prime_table(int n) {    np = 0;    for (int i = 2; i <= n; i++) {        if (vis[i]) continue;        pri[np++] = i;        for (int j = i * 2; j <= n; j += i)            vis[j] = 1;    }    vis[1] = 1;}#define lowbit(x) ((x)&(-x))struct Seg {    int l, r, id;    Seg (int l = 0, int r = 0, int id = 0) {        this->l = l;        this->r = r;        this->id = id;    }    friend bool operator < (const Seg& a, const Seg& b) {        return a.r < b.r;    }};int N, M, L[maxn], P[maxn];int fenw[maxn], ans[maxn];vector<int> g[maxn];vector<Seg> vec, que;inline void add (int x, int d) {    if (x <= 0)        return;    while (x <= N) {        fenw[x] += d;        x += lowbit(x);    }}inline int sum (int x) {    int ret = 0;    while (x) {        ret += fenw[x];        x -= lowbit(x);    }    return ret;}void init () {    vec.clear();    que.clear();    int x;    for (int i = 1; i <= N; i++) {        scanf("%d", &x);        g[i].clear();        for (int j = 0; j < np; j++) {            if (x % pri[j] == 0) {                g[i].push_back(pri[j]);                while (x % pri[j] == 0)                    x /= pri[j];            }            if (vis[x] == 0) {                g[i].push_back(x);                break;            }        }    }    for (int i = 0; i < maxn; i++) P[i] = 0;    for (int i = 1; i <= N; i++) {        int tmp = 0;        for (int j = 0; j < g[i].size(); j++) {            tmp = max(tmp, P[g[i][j]]);            P[g[i][j]] = i;        }        L[i] = tmp;        vec.push_back(Seg(tmp, i, 0));    }    for (int i = 0; i < maxn; i++) P[i] = N + 1;    for (int i = N; i; i--) {        int tmp = N + 1;        for (int j = 0; j < g[i].size(); j++) {            tmp = min(tmp, P[g[i][j]]);            P[g[i][j]] = i;        }        vec.push_back(Seg(i, tmp, 1));    }    int l, r;    for (int i = 1; i <= M; i++) {        scanf("%d%d", &l, &r);        que.push_back(Seg(l, r, i));    }}void solve () {    sort(que.begin(), que.end());    sort(vec.begin(), vec.end());    memset(fenw, 0, sizeof(fenw));    int mv = 0;    for (int i = 0; i < que.size(); i++) {        while (mv < vec.size() && vec[mv].r <= que[i].r) {            add(vec[mv].l, 1);            if (vec[mv].id)                add(L[vec[mv].l], -1);            mv++;        }        int tmp = sum(que[i].r) - sum(que[i].l - 1);        ans[que[i].id] = que[i].r - que[i].l + 1 - tmp;    }}int main () {    prime_table(maxn);    while (scanf("%d%d", &N, &M) == 2 && N + M) {        init();        solve();        for (int i = 1; i <= M; i++)            printf("%d\n", ans[i]);    }    return 0;}

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