HDU5072 Coprime【容斥原理】

传送门

There are n people standing in a line. Each of them has a unique id number.

Now the Ragnarok is coming. We should choose 3 people to defend the evil. As a group, the 3 people should be able to communicate. They are able to communicate if and only if their id numbers are pairwise coprime or pairwise not coprime. In other words, if their id numbers are a, b, c, then they can communicate if and only if [(a, b) = (b, c) = (a, c) = 1] or [(a, b) ≠ 1 and (a, c) ≠ 1 and (b, c) ≠ 1], where (x, y) denotes the greatest common divisor of x and y.

We want to know how many 3-people-groups can be chosen from the n people.

Input

The first line contains an integer T (T ≤ 5), denoting the number of the test cases.

For each test case, the first line contains an integer n(3 ≤ n ≤ 10 5), denoting the number of people. The next line contains n distinct integers a 1, a 2, . . . , a n(1 ≤ a i ≤ 10 5) separated by a single space, where a i stands for the id number of the i-th person.

Output

For each test case, output the answer in a line.

Sample Input

1
5
1 3 9 10 2

Sample Output

题意描述：给n个数，问找出其中3个数满足两两互质或者两两步互质的情况数。

思路：直接找互质不容易找，可以找出所有的不满足条件的情况。对于不满足条件的a，b，c来说必定有a，b，c三点中两点对应的两边（a点对应两边为ab，ac）的数一对数互质一对不数互质。这样题目就转化成啦求出每个数与n个数互质的个数。由之前的学习容斥我们可以求出1-n这n个数与r的互质的个数。现在是n个数，在容斥的过程中，对于一个因子积，n个数中多少与它不互质不能快速得到。但是，由于每个数都是十万以内，可以通过预处理得到。然后容斥处理一下就行啦。

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=1e5+10;
int cnt[N],a[N],pri[500],nu;
LL solve(int n)
{
    if(n==1)return LL(0);
    LL ret=-1;
    nu=0;
    for(int i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            pri[nu++]=i;
            while(n%i==0)n/=i;
        }
    }
    if(n!=1)pri[nu++]=n;
    int tim=1<<nu;
    int tem,num;
    for(int i=1;i<tim;i++)
    {
        tem=1;
        num=0;
        for(int j=0;j<nu;j++)
        {
            if(i&(1<<j))
            {
                num++;
                tem*=pri[j];
            }
        }
        if(num&1)
            ret+=cnt[tem];
        else
            ret-=cnt[tem];
    }
    return ret;
}
int main()
{
    int TA,n,x,tem;
    scanf("%d",&TA);
    while(TA--)
    {
        LL sum=0,ans,res;
        scanf("%d",&n);
        memset(cnt,0,sizeof(cnt));
        for(int i=1;i<=n;i++)scanf("%d",&a[i]);
        for(int i=1;i<=n;i++)
        {
            x=a[i];
            nu=0;
            for(int j=2;j*j<=x;j++)
            {
                if(x%j==0)
                {
                    pri[nu++]=j;
                    while(x%j==0)x/=j;
                }
            }
            if(x!=1)pri[nu++]=x;
            int tim=1<<nu;

            for(int k=1;k<tim;k++)
            {
                tem=1;
                for(int j=0;j<nu;j++)
                {
                    if(k&(1<<j))
                        tem*=pri[j];
                }
                cnt[tem]++;
            }
        }
        //for(int i=1;i<=20;i++)cout<<i<<" "<<cnt[i]<<endl;
        for(int i=1;i<=n;i++)
        {
            //cout<<"  &&";

            res=solve(a[i]);
            sum+=res*(n-res-1);
        }

        sum/=2;
        ans=(LL)n*((LL)n-1)*(n-2)/3/2;
        ans=ans-sum;
        printf("%lld\n",ans);
    }
}

HDU5072 Coprime【容斥原理】

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