题目链接：Coprime

题面：

Coprime

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

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

解题：

题意求找出三数互质或都不互质的组数，直接肯定不行。此题原型为单色三角形，结果为C（3，n）-res，其中res为每个与每个数互质和不互质数量的乘积的累加。求与每个数不互质的数量，用到了容斥原理。看似简单的原理应用却这么广泛，如果没做过容斥原理的题目，可以先试一下HDU 1796。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
bool status[100010];
int factor[100010][8];
int store[100010],cnt[100010];
long long ans[100010],res;
int one_amount[300];
int refl[300][8];
int cal[8];
void prep()
{
    memset(refl,0,sizeof(refl));
    int cont=0,temp;
   for(int i=0;i<256;i++)
   {
       temp=i;
       one_amount[i]=cont=0;
       while(temp)
       {
         if(temp%2)
         {
             one_amount[i]++;
             refl[i][cont]=1;
         }
         cont++;
         temp/=2;
       }
   }
}
bool is_prime(int a)
{
	if(a<=3)return true;
	int x=sqrt(1.0*a);
	for(int i=2;i<=x;i++)
	{
		if(a%i==0)
			return false;
	}
	return true;
}
long long C(int x,int y)
{
	long long res=1;
	for(int i=1;i<=x;i++)
	{
		res=res*(y-i+1)/i;
	}
	return res;
}
int main()
{
	int t,n,p,tmp;
	long long temp;
	scanf("%d",&t);
	prep();
	while(t--)
	{
		res=0;
		memset(status,0,sizeof(status));
		memset(factor,0,sizeof(factor));
		memset(cnt,0,sizeof(cnt));
		memset(ans,0,sizeof(ans));
		scanf("%d",&n);
		for(int i=0;i<n;i++)
			scanf("%d",&store[i]);
		for(int i=0;i<n;i++)
			status[store[i]]=1;
        for(int i=2;i<=100000;i++)
		{
	       if(is_prime(i))
		   {
			   for(int j=i;j<=100000;j+=i)
			   {
				   if(status[j])
				   {
					   cnt[i]++;
				   }
				   p=0;
				   while(factor[j][p])
				   {
					   p++;
				   }
				   factor[j][p]=i;
			   }
		   }
		   else
		   {
             for(int j=i;j<=100000;j+=i)
			   {
				   if(status[j])
				   {
					   cnt[i]++;
				   }
			   }

		   }
		}
        for(int i=0;i<n;i++)
		{
           tmp=store[i];
           p=0;
		   while(factor[tmp][p])
		   {
			   cal[p]=factor[tmp][p];
			   p++;
		   }
		   if(p==0)
		   {
			   ans[i]=0;
			   continue;
		   }
           tmp=1<<p;
		   for(int j=1;j<tmp;j++)
           {
              temp=1;
              for(int k=0;k<p;k++)
              {
                  if(refl[j][k])
                 {
                    temp=temp*cal[k];
                 }
			   }
              if(one_amount[j]%2)
              ans[i]+=cnt[temp];
              else
              ans[i]-=cnt[temp];
          }
		   ans[i]-=1;
		}
		for(int i=0;i<n;i++)
		{
			res+=(ans[i]*(n-ans[i]-1));
		}
		res/=2;
		printf("%I64d\n",C(3,n)-res);
	}
	return 0;
}

总结：

用好容斥原理的关键在于，搞清楚集合的交的含义。

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