Coprime
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 130 Accepted Submission(s): 59
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.
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, . . . , an(1 ≤ ai ≤ 105) separated by a single space, where ai stands for the id number of the i-th person.
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, . . . , an(1 ≤ ai ≤ 105) separated by a single space, where ai 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
题意：求出满足三个数的集合个数，使得每一个集合里的数满足两两互质或两两不互质
思路：因为n非常大，一定是得寻求一种nlogn的解法
注意到这题能够通过求解反面来求得答案
集合里的三个数，两两互质或两两不互质的反面是 至少有一对不互质和至少有一对互质
设这三个数为a,b,c，如今如果a与b互质，b与c不互质，设满足这种集合数位temp,那么答案为C(n,3)-temp/2，之所以除2是由于b作为中间的数算重了一次
如今问题转为求解temp
对于每一个b，设与它互质的数个数为cnt1,与它不互质的个数为cnt2，显然cnt1+cnt2+1=n
那么对于b的temp值为cnt*cnt2，当中仅仅须要求出cnt2就可以，将b分解质因子，然后这些质因子的个数不会超过6个，显然能够状压以后容斥原理求解
如今问题进而转化为求出一个数x有多少个数是x的倍数，这个用素数筛选能够求得，问题得以解决