codeforces 548 E. Mike and Foam (莫比乌斯反演)

E. Mike and Foam

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Mike is a bartender at Rico's bar. At Rico's, they put beer glasses in a special shelf. There are n kinds of beer at Rico's numbered
from 1to n. i-th
kind of beer has ai milliliters
of foam on it.



Maxim is Mike's boss. Today he told Mike to perform q queries. Initially the shelf is empty. In each request, Maxim gives him a number x.
If beer number x 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 that i < j and

where

is
the greatest common divisor of numbers aand b.

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

Input

The first line of input contains numbers n and q (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 integer x (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

思路：

问题转化一下就是给出n个数，对于某个数，和他互质的数有多少个，可以用莫比乌斯反演来解决这个问题。

对于一个数x，

f(d)表示和x的最大公约数为d的数有多少个

那么F(d)就表示和x的最大公约数为d的倍数的数有多少个

/*======================================================
# Author: whai
# Last modified: 2015-12-09 19:49
# Filename: e.cpp
======================================================*/
#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <stack>

using namespace std;

#define LL __int64
#define PB push_back
#define P pair<int, int>
#define X first
#define Y second

const int N = 5 * 1e5 + 5;

int a
;
bool used
;
int cnt
;
LL ans = 0;

int mu
;

void getMu() {
for(int i = 1; i < N; ++i) {
int tar = i == 1 ? 1 : 0;
int delta = tar - mu[i];
mu[i] = delta;
for(int j = 2 * i; j < N; j += i)
mu[j] += delta;
}
}

void add(int x) {
for(int i = 1; i * i <= x; ++i) {
if(x % i == 0) {
ans += mu[i] * cnt[i];
int tmp = x / i;
if(i != tmp)
ans += mu[tmp] * cnt[tmp];
}
}
for(int i = 1; i * i <= x; ++i) {
if(x % i == 0) {
++cnt[i];
int tmp = x / i;
if(i != tmp)
++cnt[tmp];
}
}
}

void del(int x) {
for(int i = 1; i * i <= x; ++i) {
if(x % i == 0) {
--cnt[i];
int tmp = x / i;
if(i != tmp)
--cnt[tmp];
}
}
for(int i = 1; i * i <= x; ++i) {
if(x % i == 0) {
ans -= mu[i] * cnt[i];
int tmp = x / i;
if(i != tmp)
ans -= mu[tmp] * cnt[tmp];
}
}
}

int main() {
getMu();
int n, q;
scanf("%d%d", &n, &q);
for(int i = 1; i <= n; ++i) {
scanf("%d", &a[i]);
}
for(int i = 0; i < q; ++i) {
int x;
scanf("%d", &x);
if(used[x] == 0) {
add(a[x]);
used[x] = 1;
} else {
del(a[x]);
used[x] = 0;
}
printf("%I64d\n", ans);
}
return 0;
}