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：

 1 #include<cstdio>
 2 #include<cstring>
 3 using namespace std;
 4 const int maxn=100005;
 5 int item[maxn];
 6 int  gcd(int a,int b)
 7 {
 8     if(b==0)return a;
 9     return gcd(b,a%b);
10 }
11 int main()
12 {
13   int cas,n;
14   scanf("%d",&cas);
15   while(cas--)
16   {
17     scanf("%d",&n);
18    for(int i=0;i<n;i++)
19     scanf("%d",item+i);
20    int ans=0;
21    for(int i=0;i<n;i++)
22    {
23          int numa=0,numb=0;
24         for(int j=0;j<n;j++)
25      {
26          if(i!=j)
27          {
28            if(gcd(item[i],item[j])==1)numa++;
29            else numb++;
30          }
31      }
32       ans+=numa*numb;
33    }
34     printf("%d\n",(n*(n-1)*(n-2)/6)-ans/2);
35   }
36  return 0;
37 }