HDU 5792:world is exploding (tree-like array)

http://acm.hdu.edu.cn/showproblem.php?pid=5792

World is exploding

problem DescriptionGiven a sequence A with length n,count how many quadruple (a,b,c,d) satisfies:A≠B≠C≠D,1≤A<B≤N,1≤C<D≤N,Aa<ab,aC>ad .  InputThe input consists of multiple test cases.
Each test case is begin with a integer n

The next line contains n integersa1,a2? An .
1≤n≤50000
0≤Ai≤1e9   Output For each test Case,output a line contains an integer.   Sample Input42 4 1 341 2 3 4  Sample Output 10

Insipired by http://blog.csdn.net/libin66/article/details/52098019

Test instructions: More simple and direct look.

Idea: To find out in the time before I how much smaller than it (that is, like Aa,ab, how many pairs in ascending order), with up record, behind how much smaller than it (like ac,ad such a descending number of pairs), with the down record, and then ans = up *, because there are a lot of repetition. Using a tree-like array to maintain four arrays, ls--in [1,i-1] how much smaller than a[i], rs--in [i+1,n] how much smaller than a[i], lb--in [1,i-1] how much larger than A[i, rb--in [i+1,n] there is much larger than a[i]. By the repulsion principle to subtract (LS * lb + rs * RB + ls * rs + lb * rb).

1#include <cstdio>2#include <algorithm>3#include <cstring>4 using namespacestd;5 #defineN 500056 #defineMOD 10000000077typedefLong LongLL;8 LL Ls[n], lb[n], rs[n], rb[n], bit[n];9 intA[n], num[n], N, m;Ten  One /* A ls and LB are the number of smaller numbers [1,i-1] than a[i], the number of larger numbers than A[i] - RS and RB are [I+1, n] in the number of smaller numbers than a[i], the number of larger numbers than A[i] - These can all be implemented in a tree-like array the because the answer will be counted. a=c | | a=d | | b=c | | b=d, then enumerate the four things minus it (a=c so necessarily b!=d, the same as the other - A=c:ans-=rs[i]*rb[i] - A=d:ans-=lb[i]*rb[i] - B=c:ans-=ls[i]*rs[i] + B=d:ans-=lb[i]*ls[i] - */ +  A voidInit () at { -memset (LS,0,sizeof(LS)); -memset (LB,0,sizeof(LB)); -memset (RS,0,sizeof(RS)); -Memset (RB,0,sizeof(RB)); -memset (SUM,0,sizeof(sum)); in } -  to intLowbit (intx) + { -     returnX & (-x); the } *  $ voidUpdateintPOS)Panax Notoginseng { -      while(Pos <=m) { thebit[pos]++; +pos + =Lowbit (POS); A     } the } +  -LL Query (intPOS) $ { $LL ans =0; -      while(Pos >0) { -Ans + =Bit[pos]; thepos-=Lowbit (POS); -     }Wuyi     returnans; the } -  Wu intMain () - { About      while(~SCANF ("%d", &N)) { $ init (); -          for(inti =1; I <= N; i++) { -scanf"%d", &a[i]); -Num[i] =A[i]; A         } +Sort (A +1, A +1+n); them = Unique (A +1, A +1+n)-(A +1); -          for(inti =1; I <= N; i++) $Num[i] = Lower_bound (A +1, A +1+m, Num[i])-A; the  theLL up =0, down =0; thememset (bit,0,sizeof(bit)); the          for(inti =1; I <= N; i++) { -Ls[i] = query (Num[i]-1); inLb[i] = query (M)-query (Num[i]); theup + =Ls[i]; the Update (num[i]); About         } the  thememset (bit,0,sizeof(bit)); the          for(inti = n; I >=1; i--) { +Rs[i] = query (Num[i]-1); -Rb[i] = query (M)-query (Num[i]); theDown + =Rs[i];Bayi Update (num[i]); the         } the  -LL ans =0; -  the          for(inti =1; I <= N; i++) { theAns-= ls[i] *Lb[i]; theAns-= rs[i] *Ls[i]; theAns-= rb[i] *Lb[i]; -Ans-= rb[i] *Rs[i]; the         } the  theprintf"%i64d\n", Up * down +ans);94     } the     return 0; the}

HDU 5792:world is exploding (tree-like array)