Triple acm hdu 3908)

Triple

Time Limit: 5000/3000 MS (Java/others) memory limit: 125536/65536 K (Java/Others)
Total submission (s): 387 accepted submission (s): 153

Problem descriptiongiven contains 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.

Inputthe 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.

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

Sample input1
6
2 3 5 7 11 13

Sample output20

Source2011 multi-university training contest 7-host by ECNU

Recommendxubiao

 /*  
Description:
Give you n numbers and ask how many combinations of A, B, and C meet condition 1 or condition 2.
Condition 1: Any two qualities
Condition 2: Any two do not have mutual quality.
Solution report:
You only need to count the numbers a [I]: And the number of I numbers.
And B [I]: the number of unique I numbers.
Then a [I] * B [I] is a subset of an invalid combination containing I.
It is not difficult to find that such an operation on each I can overwrite all the ABC statements that do not meet the conditions, and calculate twice.
So C (n, 3)-sum/2.

  */  
# Include  < Stdio. h  >  
  # Define  Maxn 10010  
  Int  A [maxn], B [maxn], num [maxn];
  Int  Gcd (  Int  Da,  Int  Xiao)  //  The cycle form of finding the maximum common approx.  
 {
  Int  Temp;
  While  (Xiao  ! =  0  )
{
Temp  =  Da  %  Xiao;
Da  =  Xiao;
Xiao  =  Temp;
}
  Return Da;
}
  Int  Main ()
{
  //  Freopen ("test. In", "r", stdin );
  //  Freopen ("test. Out", "W", stdout );  
    Int  T, n, I, j;
  Int  SUM;  //  Total number of nonconformities  
 Scanf (  "  % D  "  ,  &  T );
  While  (T  --  )
{
Scanf (  "  % D  "  ,  &  N );
 For  (I  =  1  ; I  <=  N; I  ++  )
{
Scanf (  "  % D  "  ,  &  Num [I]);
A [I]  =  0 ;  //  A [I] indicates the number of interchange qualities with num [I]  
  B [I]  =  0  ;  //  A [I] indicates the number of numbers that do not communicate with num [I ].  
  }
Sum  =  0  ;
  For  (I =  1  ; I  <=  N; I  ++  )
{
  For  (J  =  1  ; J  <  I; j  ++  )
{
  If (Gcd (Num [I], num [J])  =  1  ) A [I]  ++  ;
  Else  B [I]  ++  ;
}
  For  (J  =  I  +  1  ; J <=  N; j  ++  )
{
  If  (Gcd (Num [I], num [J])  =  1  ) A [I]  ++  ;
  Else  B [I]  ++  ;
}
Sum  + =  A [I] *  B [I];
}
  Int  CNT  =  N  *  (N  -  1  )  *  (N  -  2  )  /  6 -  Sum  /  2  ;
Printf (  "  % D \ n  "  , CNT );
}
  Return     0  ;
}