Poj 2299 Ultra-QuickSort (tree array/Reverse Order Number)

Poj 2299 Ultra-QuickSort (tree array/Reverse Order Number)
Ultra-QuickSort

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

Description

In this problem, you have to analyze a particle sorting algorithm. the algorithm processes a sequence of n distinct integers by swapping two 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 how swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.

Input

The input contains several test cases. every test case begins with a line that 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 not be processed.

Output

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

Sample Input

59105431230

Sample Output

60

Source

Waterloo local 2005.02.05

#include 
 
  #include 
  
   #include 
   
    #include
    
     #include #include
     
      using namespace std;int b[500005], c[500005];int n;struct node{    int num, id;} a[500005];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;}