HDU-4777 Rabbit Kingdomom(树状数组、区间离线)

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title: HDU-4777 Rabbit Kingdomom(树状数组、区间离线)
date: 2018-12-14 11:20:15
tags: [树状数组,离线,区间]
categories: ACM

题目链接

Problem Description

Long long ago, there was an ancient rabbit kingdom in the forest. Every rabbit in this kingdom was not cute but totally pugnacious, so the kingdom was in chaos in season and out of season.
　　n rabbits were numbered form 1 to n. All rabbits’ weight is an integer. For some unknown reason, two rabbits would fight each other if and only if their weight is NOT co-prime.
　　Now the king had arranged the n rabbits in a line ordered by their numbers. The king planned to send some rabbits into prison. He wanted to know that, if he sent all rabbits between the i-th one and the j-th one(including the i-th one and the j-th one) into prison, how many rabbits in the prison would not fight with others.
　　Please note that a rabbit would not fight with himself.

Input

The input consists of several test cases.
　　The first line of each test case contains two integer n, m, indicating the number of rabbits and the queries.
　　The following line contains n integers, and the i-th integer Wi indicates the weight of the i-th rabbit.
　　Then m lines follow. Each line represents a query. It contains two integers L and R, meaning the king wanted to ask about the situation that if he sent all rabbits from the L-th one to the R-th one into prison.
　　(1 <= n, m, Wi <= 200000, 1 <= L <= R <= n)
　　The input ends with n = 0 and m = 0.

Output

For every query, output one line indicating the answer.

Sample Input

3 2
2 1 4
1 2
1 3
6 4
3 6 1 2 5 3
1 3
4 6
4 4
2 6
0 0

Sample Output

2
1
1
3
1
2

Hint

In the second case, the answer of the 4-th query is 2, because only 1 and 5 is co-prime with other numbers in the interval [2,6] .

AC

• 询问与区间内数字互质的个数
• 可以统计每个数字三个区间（左边最近不互质点，i），（i, 右边最近不互质点），（左边最近不互质，右边最近不互质）
• ans = 区间长度 - 区间1 - 区间2 + 区间3

#include <iostream>
#include <stdio.h>
#include <map>
#include <unordered_map>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#include <cstring>
#include <cmath>
#include <iomanip>
#include <algorithm>
#define N 200005
#define lowbit(x) (x & (-x))
#define mem(a, b) memset(a, b, sizeof(a))
#define REP(i, n) for (int i = 1; i <= (n); ++i)
#define rep(i, n) for (int i = 0; i < (n); ++i)
typedef long long LL;
using namespace std;
int n, m, a[N], L[N], R[N], ans[3][N], c[N], len[N];
struct ac{
    int l, r, id;
    bool operator <(const ac &t) const{
        return r < t.r;
    }
}q[N], sum[3][N];

void init() {
    map<int, int> mp;
    mp.clear();
    REP(i, n) {
        L[i] = 0;
        int t = a[i];
        for (int j = 2; j * j <= t; ++j) {
            if (t % j)  continue;
            if(mp.find(j) != mp.end())  L[i] = max(L[i], mp[j]);
            mp[j] = i;
            while (t % j == 0)  t /= j;
        }
        if (t > 1)  {
            if (mp.find(t) != mp.end()) L[i] = max(L[i], mp[t]);
            mp[t] = i;
        }
    }
    mp.clear();
    for (int i = n; i >= 1; --i) {
        int t = a[i];
        R[i] = n + 1;
        for (int j = 2; j * j <= t; ++j) {
            if (t % j)  continue;
            if (mp.find(j) != mp.end()) R[i] = min(R[i], mp[j]);
            mp[j] = i;
            while (t % j == 0)  t /= j;
        }
        if (t > 1)  {
            if (mp.find(t) != mp.end()) R[i] = min(R[i], mp[t]);
            mp[t] = i;
        }
    }
    REP(i, n) {
        sum[0][i] = {L[i], i};  // 记录点左边不互质区间
        sum[1][i] = {i, R[i]};  // 记录点右边不互质区间
        sum[2][i] = {L[i], R[i]};   // 记录点左右不互质区间
    }
    // 按照有区间排序
    sort (sum[0] + 1, sum[0] + n + 1);
    sort (sum[1] + 1, sum[1] + n + 1);
    sort (sum[2] + 1, sum[2] + n + 1);
}
void update(int x) {
    if (x == 0) return; // 左边不存在不互质的点
    while (x <= n) {
        c[x] += 1;
        x += lowbit(x);
    }
}

int Sum(int x) {
    int sum = 0;
    while (x > 0) {
        sum += c[x];
        x -= lowbit(x);
    }
    return sum;
}

void solve() {
    rep(i, 3) {
        mem(c, 0);  // 清空树状数组
        int j = 1, k = 1;
        while (k <= m) {
            while (j <= n && sum[i][j].r <= q[k].r) {
                update(sum[i][j].l);
                ++j;
            }
            ans[i][q[k].id] = Sum(q[k].r) - Sum(q[k].l - 1);
            ++k;
        }
    }
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
#endif
    while (scanf("%d %d", &n, &m) != EOF) {
        if (n + m == 0) break;
        REP(i, n) scanf("%d", &a[i]);
        REP(i, m) {
            scanf("%d %d", &q[i].l, &q[i].r);
            q[i].id = i;
            len[i] = q[i].r - q[i].l + 1;
        }
        sort(q + 1, q + m + 1);
        init();
        solve();
        REP(i, m) {
            printf("%d\n", len[i] - ans[0][i] - ans[1][i] + ans[2][i]);
        }
    }

    return 0;
}