HDU 4000 Fruit Ninja

/*  Question: Given the sequence, one of the subsequences is (x, y, z). Ask how many of these subsequences are there at most, so that x <z <y, in this analysis, the sequence is 1 to n: because it is an arrangement, so there is no same element, so x, y, z are not equal to each other. Therefore, the size relationship in the sequence can be expressed as small, medium, large, and small, and = small ____, so the answer is small, medium, and small _-small, medium, and small _. You can count the number of all elements after this bit as X, then the small _ = x * (x-1)/2 and small statistics in the big, just need to count the first and later than his big number x, y, the number of small-to-large = x * y statistics can be calculated using a tree array. The tree array can calculate the number of previously added elements smaller than the current element.  */  # Include <Iostream> # Include <Cstdio> # Include <Cstring> Using   Namespace  STD;  # Define Mod4 100000007 # Define X 100005 Int  A [x];  Long   Long  C [X], big [X], SMA [x];  Int Lowbit ( Int  X ){  Return X &- X ;}  Void Modify ( Int  X ){  While (X <X) {C [x] = (C [x] + 1 ) % MOD; x + = Lowbit (x );}}  Long   Long Query ( Int  X ){  Long   Long Ret = 0  ;  While (X> 0  ) {RET + = C [X]; RET % = MOD; x -= Lowbit (x );}  Return  RET ;}  Int  Main () {freopen (  "  Sum. In  " , "  R  "  , Stdin); freopen (  " Sum. Out  " , "  W  "  , Stdout );  Int CNT = 0  , T, n; CIN > T;  While (T -- ) {Printf (  "  Case # % d:  " , ++CNT); scanf (  "  % D  " ,& N); memset (C,  0 , Sizeof  (C); memset (big,  0 , Sizeof  (BIG); memset (SMA,  0 , Sizeof  (SMA ));  For (Int I = 1 ; I <= N; I ++ ) {Scanf (  "  % D  " ,& A [I]); SMA [I] = Query (A [I]); ///  Count the number smaller than him  Modify (A [I]);} memset (C,  0 , Sizeof  (C ));  For (Int I = N; I> = 1 ; I --) ///  Count the number later than him  {Big [I] = N-I- Query (A [I]); Modify (A [I]);}  Long   Long S = 0 , B = 0  ;  For ( Int I = 1 ; I <= N; I ++ ) {S = (S + (BIG [I]- 1 ) * Big [I]/ 2 ) % MOD; ///  Count the number of small _ B = (B + big [I] * SMA [I]) % MOD; ///  Count the numbers in a Small and Medium Size  } Cout <(S-B + mod) % mod < Endl ;}  Return   0  ;}