HDU Coprime

Coprime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 433    Accepted Submission(s): 192

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

题意：求n个数字中取出3个数字，统计   都互质  或  都不互质  的对数。

思路：取出的3个数字a,b,c 有8种情况。

其中题目要求的只是其中两种情况，还有6种情况，要么就是1对互质对，要么有2队互质对存在。

如果能求得 ai 在n个数字中 有 k 个数字与其不互质   k2个数字与其互质。

那么题目就能转化为求所有的   sum（k*k2）;

对于数字ai来说，我们知道已经有一对互质对必然存在了，但是也可能会存在第2对，因为k个中和k2中。

所以，就相当于对ai来说，有至少一对互质对，最多2个互质对的情况。 统计所有。

我们会发现，会多算了一次，/2即可。

下面就是关于如何求ai 有多少个数字与ai互质的问题了。

容斥。

 #include<iostream>
 #include<stdio.h>
 #include<cstring>
 #include<cstdlib>
 #include<vector>
 using namespace std;
 const int maxn = 1e5+;

 int a[maxn];
 bool s[maxn];
 int prime[maxn],len;
 vector<int>Q[maxn];
 vector<int>X[maxn];
 int Hash[maxn];

 int H[maxn],hlen;
 void init()
 {
     memset(s,false,sizeof(s));
     s[]=true;
     len = ;
     for(int i=; i<maxn; i++)
     {
         if(s[i]==true)continue;
         prime[++len]=i;
         for(int j=i+i; j<maxn; j=j+i)
             s[j] = true;
     }
     /**筛选i素因子**/
     for(int i=; i<=len; i++)
     {
         for(int j=prime[i]; j<maxn; j=j+prime[i])
             Q[j].push_back(prime[i]);
     }
     int k,tmp;
     for(int i=; i<maxn; i++){
         hlen = ;
         H[]=-;
         k = Q[i].size();
         for(int j=; j<k; j++){
             tmp = hlen;
             for(int ss=; ss<=tmp; ss++){
                 H[++hlen]=-*H[ss]*Q[i][j];
                 X[i].push_back(H[hlen]);
             }
         }
     }
 }
 int main()
 {
     int T,n,k;
     init();
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d",&n);
         memset(Hash,,sizeof(Hash));
         for(int i=; i<=n; i++)
         {
             scanf("%d",&a[i]);
             k = X[a[i]].size();
             for(int j=; j<k; j++)
                 if(X[a[i]][j]<)
                     Hash[-X[a[i]][j]]++;
                 else
                     Hash[X[a[i]][j]]++;
         }
         int sum;
         __int64 tom = ;
         for(int i=; i<=n; i++)
         {
             if(a[i]==) continue;
             sum = ;
             k = X[a[i]].size();
             /**sum 与ai 不互质的个数**/
             for(int j=; j<k; j++)
             {
                 if(X[a[i]][j]<)
                     sum=sum-Hash[-X[a[i]][j]];
                 else sum=sum+Hash[X[a[i]][j]];
             }
             sum --; //减去本身
             tom = tom + sum*(__int64)(n--sum);
         }
         printf("%I64d\n",(n*(__int64)(n-)*(n-))/-tom/);
     }
     return ;
 }