CodeForces 548E Mike and Foam （容斥+数论）

E. 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

先预处理出一个数的素因子，和已经存在的数的质因子做容斥·

#include "cstring"#include "cstdio"#include "iostream"#include "string.h"#include "vector"#include "cmath"#define LL long longusing namespace std;int a[200005];int vis[200005];int cnt[500005];long long total,ans,len;//long long s[500005];vector <long long> list[500005];void makeList(long long i){    long long old=i;    for(long long j=2;j<=sqrt(i);j++)    {        if(i%j==0)        {            list[old].push_back(j);            while(i%j==0)                i/=j;        }    }    if(i!=1)        list[old].push_back(i);}long long add(int x){    vector <long long> now=list[x];    long long len=list[x].size();    long long sum=0;    for(int i=1;i<(1<<len);i++)    {        long long id=1;        long long flag=0;        for(int j=0;j<len;j++)        {            if(i&(1<<j))            {                id*=now[j];                //id*=s[j];                flag++;            }        }        if(flag%2==0)            sum-=cnt[id];        else            sum+=cnt[id];        cnt[id]++;    }    ans+=(total-sum);    total++;    return ans;}long long sub(int x){    vector <long long> now=list[x];    long long len=list[x].size();    long long sum=0;    for(int i=1;i<(1<<len);i++)    {        long long id=1;        long long flag=0;        for(int j=0;j<len;j++)        {            if(i&(1<<j))            {                id*=now[j];                flag++;            }        }        cnt[id]--;        if(flag%2==0)            sum-=cnt[id];        else            sum+=cnt[id];    }    total--;    ans-=(total-sum);    return ans;}int main(){    int n,q;    int mark[500005];    memset(mark,0,sizeof(mark));    while(~scanf("%d%d",&n,&q))    {        memset(vis,0,sizeof(vis));        memset(cnt,0,sizeof(cnt));        total=0;        ans=0;        for(int i=1;i<=n;i++)            scanf("%d",&a[i]);        while(q--)        {            int now;            scanf("%d",&now);            if(mark[a[now]]==0)            {                makeList(a[now]);                mark[a[now]]=1;            }            //_set(a[now]);            if(vis[now]==0)            {                vis[now]=1;                printf("%lld\n",add(a[now]));            }            else            {                vis[now]=0;                printf("%lld\n",sub(a[now]));            }        }    }}