7_11_ H题 Rabbit Kingdom（容斥+树状数组）

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题意

给出一个序列以及若干询问，询问是一个区间[l,r],问给出的区间中和其他数都互质的数的个数。

思路

对每个数先处理出他左右最近的和他不互质的数的位置记为L[x]，R[x]，这样问题就变成了求区间[l,r]中有多少个数的L和R都不在[L,R]内，简单推一下，再容斥一下就可以得到下面的结果

S=R−L+1SLorR=sum(L[x]在[L,R]内)+sum(R[x]在[L,R]内)SLR=sum（L[x]，R[x]都在[L,R]内）ans=S−(SLorR−SLR)

求左右区间在不在目标区间内可以用左右端点标记+1-1再求和的方式处理。这里要离线处理，用树状数组求和。

代码

#include <bits/stdc++.h>using namespace std;const int maxn = 2e5+10;struct Node{    int l, r, idx;    Node (int l = 0, int r = 0, int idx = 0):l(l),r(r),idx(idx){}    friend bool operator < (const Node& a, const Node& b) {        return a.r < b.r;    }};int pri[maxn], vis[maxn];int cnt ,N, M, L[maxn], P[maxn];int  ans[maxn];vector<int> fact[maxn];vector<Node> vec, qur;void getprime(int n) {    cnt = 0;    for (int i = 2; i <= n; i++) {        if (vis[i]) continue;        pri[cnt++] = i;        for (int j = i * 2; j <= n; j += i) vis[j] = 1;    }    vis[1] = 1;}/*------------------树状数组---------------------*/#define lowbit(x) ((x)&(-x))int tree[maxn];inline void add (int x, int d) {    if (x <= 0) return;    while (x <= N) {        tree[x] += d;        x += lowbit(x);    }}inline int sum (int x) {    int ret = 0;    while (x) {        ret += tree[x];        x -= lowbit(x);    }    return ret;}/*--------------------------------------*/void init () {    vec.clear();    qur.clear();    memset(tree, 0, sizeof tree);    for (int i = 1; i <= N; i++) {        int x;        scanf("%d", &x);        fact[i].clear();        for (int j = 0; j < cnt; j++) {            if (x % pri[j] == 0) {                fact[i].push_back(pri[j]);                while (!(x % pri[j]))                    x /= pri[j];            }            if (!vis[x]) {                fact[i].push_back(x);                break;            }        }    }}void dealLR(){    for (int i = 0; i < maxn; i++) {        P[i] = 0;    }    for (int i = 1; i <= N; i++) {        int l = 0;        for (int j = 0; j < fact[i].size(); j++) {            l = max(l, P[fact[i][j]]);            P[fact[i][j]] = i;        }        L[i] = l;        vec.push_back(Node(l, i, 0));    }    for (int i = 0; i < maxn; i++) {        P[i] = N + 1;    }    for (int i = N; i > 0; i--) {        int r = N + 1;        for (int j = 0; j < fact[i].size(); j++) {            r = min(r, P[fact[i][j]]);            P[fact[i][j]] = i;        }        vec.push_back(Node(i, r, 1));    }}int main () {    getprime(maxn);    while (~scanf("%d%d", &N, &M)&& N) {        init();        dealLR();        for (int i = 1; i <= M; i++) {            int l, r;            scanf("%d %d", &l, &r);            qur.push_back(Node(l, r, i));        }        sort(qur.begin(), qur.end());        sort(vec.begin(), vec.end());        int k = 0;        for (int i = 0; i < qur.size(); i++) {            for(; k < vec.size() && vec[k].r <= qur[i].r ; k++){                add(vec[k].l, 1);                if (vec[k].idx)                    add(L[vec[k].l], -1);            }            int su = sum(qur[i].r) - sum(qur[i].l-1);            ans[qur[i].idx] = qur[i].r - qur[i].l - su + 1;        }        for (int i = 1; i <= M; i++) {            printf("%d\n", ans[i]);        }    }    return 0;}