POJ 2299 ultra-quicksort (tree array/reverse order number)

Ultra-quicksort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 46080		Accepted: 16763

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

#include <iostream> #include <stdio.h> #include <cmath> #include <stdlib.h> #include < Algorithm> #include <string.h>using namespace Std;int b[500005], c[500005];int n;struct node{int num, id;} a[50 0005];bool CMP (Node A, Node B) {return a.num < B.num;} int lowbit (int x) {return x& (-X);}        void Update (int i, int x) {while (I <= N) {C[i] + = x;    i = i + lowbit (i);    }}int sum (int i) {int sum = 0;        while (i > 0) {sum + = C[i];    i = I-lowbit (i); } return sum;}    int main () {int i;    Long long ans;        while (scanf ("%d", &n)!=eof) {memset (b, 0, sizeof (b));        memset (c, 0, sizeof (c));            for (i = 1; I <= n; i++) {scanf ("%d", &a[i].num);        A[i].id = i;        } sort (a+1, a+n+1, CMP);        B[a[1].id] = 1;        for (i = 2; I <= n; i++) {b[a[i].id] = i;        } ans = 0;      for (i = 1; I <= n; i++)  {Update (b[i], 1);        Ans + = (sum (n)-sum (b[i));    } printf ("%lld\n", ans); } return 0;}

POJ 2299 ultra-quicksort (tree array/reverse order number)