【codeforces 548E】Mike and Foam

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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 1 to 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 a and 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.

Examples
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

【题解】

gcd(x,y)==1就表示这两个数是互质的;
如何求一些数中与某个数互质的个数,参考这篇文章http://blog.csdn.net/harlow_cheng/article/details/53870341
如果新加入的数没有在架子上
则在架子上的数字里面找和这个数不互质的数;
然后用总数减去它，就是与这个数互质的数的个数了；
然后答案加上这个个数.
在做这个的过程中记录某个数的倍数的个数num;
这样能直接累加不互质的数的个数;
不理解方法的话多看那个链接里面的例题就好.
如果在架子上，就先把这个数取出来;然后再找和它互质的数的个数;
然后答案减去这个个数;

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int MAXN = 5e5+10;
const int MAXN2 = 2e5+10;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);

int num[MAXN],n,q,pri[200],len,total,a[MAXN2];
bool bo[MAXN2];
LL ans = 0;

void get_pri(int x)
{
    len = 0;
    for (int i = 2;i*i <= x;i++)
        if ((x%i)==0)
        {
            pri[++len] = i;
            while ((x%i)==0)
                x/=i;
        }
    if (x > 1)
        pri[++len] = x;
}

void sub()
{
    LL tot = 0;
    for (int i = 1;i <= (1<<len)-1;i++)
    {
        LL sum = 1,f = -1;
        rep1(j,1,len)
            if (i&(1<<(j-1)))
            {
                sum*=pri[j];
                f*=-1;
            }
        num[sum]--;
        tot += f*num[sum];
    }
    total--;
    ans -= (total-tot);
}

void add()
{
    LL tot = 0;
    for (int i = 1;i <= (1<<len)-1;i++)
    {
        LL sum = 1,f = -1;
        rep1(j,1,len)
            if (i&(1<<(j-1)))
            {
                sum*=pri[j];
                f*=-1;
            }
        tot += f*num[sum];
        num[sum]++;
    }
    ans += (total-tot);
    total++;
}

int main()
{
    //freopen("F:\rush.txt","r",stdin);
    rei(n);rei(q);
    rep1(i,1,n)
        rei(a[i]);
    rep1(i,1,q)
        {
            int x;
            rei(x);
            get_pri(a[x]);
            if (bo[x])
            {
                sub();
                bo[x] = false;
            }
            else
            {
                add();
                bo[x] = true;
            }
            printf("%I64d
",ans);
        }
    return 0;
}