Poj2299:ultra-quicksort (tree-like array for reverse order number)

Description

In this problem, you has to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping, adjacent sequence elements until the sequence is Sorted in ascending order. For the input sequence
9 1 0 5 4,
Ultra-quicksort produces the output
0 1 4 5 9.
Your task is to determine what many swap operations Ultra-quicksort needs to perform in order to sort a given input sequenc E.

Input

The input contains several test cases. Every test case begins with a line this contains a single integer n < 500,000-the length of the input sequence. Each of the following n lines contains a single integer 0≤a[i]≤999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must is processed.

Output

For every input sequence, your program prints a single line containing an integer number OP, the minimum number of swap op Erations necessary to sort the given input sequence.

Sample Input

59105431230

Sample Output

60

Source

Waterloo Local 2005.02.05
Test instructions: To sort by exchange, to seek the number of exchanges, in fact, the number of reverse order
Idea: Using a tree-shaped array is a method, for a large value, we need to first use discretization

#include <iostream> #include <stdio.h> #include <string.h> #include <string> #include <stack > #include <queue> #include <map> #include <set> #include <vector> #include <math.h># Include <bitset> #include <list> #include <algorithm> #include <climits>using namespace std;# Define Lson 2*i#define Rson 2*i+1#define LS l,mid,lson#define RS mid+1,r,rson#define Up (i,x,y) for (i=x;i<=y;i++) # Define down (i,x,y) for (i=x;i>=y;i--) #define MEM (a,x) memset (A,x,sizeof (a)) #define W (a) while (a) #define GCD (A, B) __ GCD (A, b) #define LL long long#define N 500005#define INF 0x3f3f3f3f#define EXP 1e-8#define lowbit (x) (x&-x) const int MO D = 1e9+7;struct node{int x,id;}    A[n];int N,c[n],r[n];int CMP (node A,node b) {if (a.x!=b.x) return a.x<b.x; return a.id<b.id;}    int sum (int x) {int ret=0;        while (x>0) {ret+=c[x];    X-=lowbit (x); } return ret;}    void Add (int x,int d) {while (x<=n) {    C[x]+=d;    X+=lowbit (x);    }}int Main () {int i,j,k;            while (~SCANF ("%d", &n), N) {for (i = 1;i<=n;i++) {scanf ("%d", &a[i].x);        A[i].id = i;        } sort (a+1,a+1+n,cmp);        MEM (c,0);        for (i = 1;i<=n;i++) {R[a[i].id] = i;        } LL ans = 0;            for (i = 1;i<=n;i++) {Add (r[i],1);//insert in order of input, first for the currently entered value, the statistic is not smaller than its value has been inserted, and then I minus the data, you can get the reverse number        ans+= (I-sum (r[i));    } printf ("%i64d\n", ans); } return 0;}

Poj2299:ultra-quicksort (tree-like array for reverse order number)