hdu 5072 Coprime

Coprime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 56    Accepted Submission(s): 20

Problem Description

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 ≤ 10⁵), denoting the number of people. The next line contains n distinct integers a₁, a₂, . . . , a_n(1 ≤ a_i ≤ 10⁵) 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

4

 

Source

2014 Asia AnShan Regional Contest 

 

先用容斥求出和a[i] 互质的个数num ，然后不符合条件的 就是 num*(n-1-num);

把所有不符合的加起来/2， 就是所有不符合的组数，最后用总的组数减去不合法的就好了

对于 num 的求法：

 先用cnt[i] 表示。因子里面有 i 的数的个数。 

对于a[i] ,先分解质因子，然后就可以用容斥求了，具体看代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<set>
#include<stack>
#include<map>
#include<ctime>
#include<bitset>
#define LL long long
#define INF 999999
#define maxn 500010
using namespace std;

int cnt[maxn] ,a[maxn] ;
vector<int>q;
int main()
{
    int i ,j , n ,m , k , v ;
    LL ans,sum;
    int T ,len,u;
    cin >>T ;
    while(T--)
    {
        scanf("%d",&n) ;
        for( i =1 ; i <= n ;i++)
            scanf("%d",&a[i]) ;
        memset(cnt,0,sizeof(cnt)) ;
        for( i = 1 ; i <= n ;i++)
        {
            for( j = 1 ; j*j <= a[i] ;j++)if(a[i]%j==0)
            {
                cnt[j]++ ;
                if(a[i]/j != j )cnt[a[i]/j]++;
            }
        }
        ans = 0 ;
        for( i = 1 ; i <= n ;i++)
        {
            q.clear();
            m = a[i];
            for( j = 2 ; j*j <= m ;j++)if(m%j==0)
            {
                q.push_back(j) ;
                while(m%j==0) m /= j ;
            }
            if(m != 1) q.push_back(m) ;
            k = q.size() ;
            len = (1<<k) ;
            int hehe=0;
            sum=0;
            u=1;
            for( j = 1 ;j < len ;j++)
            {
                hehe=0;
                u=1;
                for( v = 0 ; v < k ;v++ )
                    if((1<<v)&j)
                {
                    hehe++ ;
                    u *= q[v] ;
                }
                if(hehe&1) sum += cnt[u] ;
                else sum -= cnt[u] ;
            }
            if(sum)sum--;
            ans += sum*(n-1-sum) ;
        }
        ans/=2;
        LL hehe = (LL)n*(n-1)*(n-2)/6 ;
        printf("%I64d
",hehe-ans) ;
    }
    return 0 ;
}