Codeforces 548E Mike and Foam (容斥＋莫比乌斯反演)

C. Mike and Foam
time limit per test：2 seconds

memory limit per test：256 megabytes

Mike is a bartender at Rico's bar. At Rico's, they put beer glasses in a special shelf. There aren kinds of beer at Rico's numbered from1 to
n. i-th kind of beer hasai milliliters of foam on it.



Maxim is Mike's boss. Today he told Mike to performq queries. Initially the shelf is empty. In each request, Maxim gives him a numberx. If beer numberx
is already in the shelf, then Mike should remove it from the shelf, otherwise he should put it in the shelf.

After each query, Mike should tell him the score of the shelf. Bears are geeks. So they think that the score of a shelf is the number of pairs(i, j) of glasses in the shelf such
thati < j and

where


is the greatest common divisor of numbersa andb.

Mike is tired. So he asked you to help him in performing these requests.

Input
The first line of input contains numbers
n andq (1 ≤ n, q ≤ 2 × 105), the number of different kinds of beer and number of queries.

The next line contains n space separated integers,a1, a2, ... , an
(1 ≤ ai ≤ 5 × 105), the height of foam in top of each kind of beer.

The next q lines contain the queries. Each query consists of a single integer integerx (1 ≤ x ≤ n),
the index of a beer that should be added or removed from the shelf.

Output
For each query, print the answer for that query in one line.

Sample test(s)

Input

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

Output

0
1
3
5
6
2

题目链接：http://codeforces.com/contest/548/problem/E

题目大意：n个数，q次询问，每次询问给出一个下标，如果询问位置未出现，则将数字加入容器，否则从容器中删除对应位置的元素，查询的是容器中gcd(ai, aj) = 1的对数(i < j)

题目分析：典型的容斥题，预处理出每个数字的约数，设F(x)为gcd(ai, aj)＝x的倍数时的个数，G(x)为gcd(ai, aj) = x时的倍数

则有F(x) = ΣG(kx) (k >= 1)，莫比乌斯反演得

G(x) = ΣF(kx)u(k) => G(1) = ΣF(k)u(k)

每次操作时维护一个cnt数组，cnt[i]表示i的倍数的个数，这样就不用计算F(k)了，因为1个数的时候是0，加的时候先计算当前的再加，减的时候先减再计算，注意每次计算的是ans的累计量

比如样例：

1：cnt[1] = 0

2：cnt[1] = 1，cnt[2] = 0

3：cnt[1] = 2，cnt[3] = 0

4：cnt[1] = 3，cnt[2] = 1，cnt[4] = 0

5：cnt[1] = 4，cnt[2] = 2，cnt[3] = 1，cnt[6] = 0 可以发现Σcnt[1] = 10 = C(4,2)

1：cnt[1] = 4

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#define ll long long
using namespace std;
int const MAX = 5e5 + 5;
int p[MAX], mob[MAX], a[MAX], cnt[MAX];
bool noprime[MAX], vis[MAX], has[MAX];
vector <int> vt[MAX];

void Mobius()
{
    int pnum = 0;
    mob[1] = 1;
    for(int i = 2; i < MAX; i++)
    {
        if(!noprime[i])
        {
            p[pnum ++] = i;
            mob[i] = -1;
        }
        for(int j = 0; j < pnum && i * p[j] < MAX; j++)
        {
            noprime[i * p[j]] = true;
            if(i % p[j] == 0)
            {
                mob[i * p[j]] = 0;
                break;
            }
            mob[i * p[j]] = -mob[i];
        }
    }
}

int main()
{
    Mobius();
    int n, q, idx, ma = 0;
    scanf("%d %d", &n, &q);
    for(int i = 1; i <= n; i++)
    {
        scanf("%d", &a[i]);
        has[a[i]] = true;
        ma = max(ma, a[i]);
    }
    for(int i = 1; i <= ma; i++)
        for(int j = i; j <= ma; j += i)
            if(has[j])
                vt[j].push_back(i);
    ll ans = 0;
    while(q--)
    {
        scanf("%d", &idx);
        if(!vis[idx])
        {
            vis[idx] = true;
            for(int i = 0; i < (int) vt[a[idx]].size(); i++)
            {
                int num = vt[a[idx]][i];
                ans += (ll) mob[num] * cnt[num];
                cnt[num] ++;
            }
        }
        else
        {
            vis[idx] = false;
            for(int i = 0; i < (int) vt[a[idx]].size(); i++)
            {
                int num = vt[a[idx]][i];
                cnt[num] --;
                ans -= (ll) mob[num] * cnt[num];
            }
        }
        printf("%I64d\n", ans);
    }
}