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 < 10 ⁵) 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 }

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