hdu 3908 Triple(组合计数、容斥原理)

Triple

Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others) Total Submission(s): 1365    Accepted Submission(s): 549

Problem Description

Given many different integers, find out the number of triples (a, b, c) which satisfy a, b, c are co-primed each other or are not co-primed each other. In a triple, (a, b, c) and (b, a, c) are considered as same triple.

 

Input

The first line contains a single integer T (T <= 15), indicating the number of test cases. In each case, the first line contains one integer n (3 <= n <= 800), second line contains n different integers d (2 <= d < 10⁵) separated with space.

 

Output

For each test case, output an integer in one line, indicating the number of triples.

 

Sample Input

1 6 2 3 5 7 11 13

 

Sample Output

20

 

Source

2011 Multi-University Training Contest 7 - Host by ECNU

 

给你n个数，对于其中的任意n个数，a,b,c 要么两两互斥，要么a,b,c两两不互斥......

要你求出满足这一条件的组合数。

分析：

    对于任意的三个数，a，b,c 我们知道有这些情况，0对互斥（即两两都不互斥），1对互斥，两对互斥，三对互斥（即两两互斥）。

  其后思路与其相同： http://blog.csdn.net/cool_fires/article/details/8681888

  代码：

 

#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=100005;
int item[maxn];
int  gcd(int a,int b)
{
    if(b==0)return a;
    return gcd(b,a%b);
}
int main()
{
  int cas,n;
  scanf("%d",&cas);
  while(cas--)
  {
    scanf("%d",&n);
   for(int i=0;i<n;i++)
    scanf("%d",item+i);
   int ans=0;
   for(int i=0;i<n;i++)
   {
         int numa=0,numb=0;
        for(int j=0;j<n;j++)
     {
         if(i!=j)
         {
           if(gcd(item[i],item[j])==1)numa++;
           else numb++;
         }
     }
      ans+=numa*numb;
   }
    printf("%d
",(n*(n-1)*(n-2)/6)-ans/2);
  }
 return 0;
}