Codevs 4163 The number of reverse pairs-tree Array method

4163 Hzwer and reverse ordertime limit: ten sspace limit: 256000 KBtitle level: Golden GoldTitle Description Description

Hzwer in the study of reverse order.

For the sequence {a}, if the ordinal pair (i,j) satisfies: I<j,a[i]>a[j], then (I,J) is a pair of inverse pairs.

Given a sequence {a}, the number of inverse pairs is obtained.

Input data is large, please use scanf instead of CIN to read in.

Enter a description Input Description

The first row is a number n, which indicates that {a} has n elements.

Next n number, describe {a}.

Output description Output Description

A number that represents the number of reverse orders.

Sample input Sample Input

5

3 1 5) 2 4

Sample output Sample Output

4

Data range and Tips Data Size & Hint

For 10% data, 1<=n<=100.

For 20% data, 1<=n<=10000.

For 30% data, 1<=n<=100000.

For 100% data, 1<=n<=1000000,1<=a[i]<=10^8.

#include <cstdio>#include<iostream>using namespacestd; #include<algorithm>#include<cstring>#defineMAXN 1000001structNode {Long Longv; intID; BOOL operator< (ConstNode &p)Const{returnv<p.v;}}; Node A[MAXN+Ten];Long Long intc[maxn+Ten],b[maxn+Ten];intN;inlineintLowbit (intx) {    returnx& (-x);}Long LongQueryintx) {    Long Longans=0;  while(x) {ans+=C[x]; X-=lowbit (x); }    returnans;}voidChangeintx) {     while(x<=N) {c[x]++; X+=lowbit (x); }}intMain () {scanf ("%d",&N); Memset (A,0,sizeof(a)); Memset (c,0,sizeof(c)); memset (b,0,sizeof(b));  for(intI=1; i<=n;++i) {scanf ("%d",&(A[I].V)); A[i].id=i; } sort (A+1, a+n+1); intpre=-1; intPrevalue=0;  for(intI=1; i<=n;++i) {if(pre!=a[i].v) {Pre=a[i].v; A[I].V=++Prevalue; }        Elsea[i].v=Prevalue; }     for(intI=1; i<=n;++i) b[a[i].id]=a[i].v; Long Long ints=0;  for(inti=n;i>=1;--i) {change (b[i]); S+=query (b[i]-1); } cout<<s<<Endl; return 0;}

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Codevs 4163 The number of reverse pairs-tree Array method