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Problem F

Ultra-QuickSort _p

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

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Description

Problem B: Ultra-QuickSort

In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of ndistinct 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 many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.

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 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.

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

5
9
1
0
5
4
3
1
2
3
0

Output for Sample Input

6
0

---

Stefan B�ttcher

Input

Output

Sample Input

Sample Output

Hint

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Problem descriptions:

System Crawler 2011-08-31 0 0

Initialization.

---

#include <cstdio>
#include <vector>
#include <cstring>
#include <algorithm>
using namespace std;
#define LOW(x) ((x)&(-x))
int n;
int c[500010];
void mod ( int x, int v ) {
 while ( x <= n ) {
  c[x] += v;
  x += LOW ( x ); 
 } 
}
long long sum ( int x ) {
 long long s = 0;
 while ( x > 0 ) {
  s += c[x];
  x -= LOW ( x ); 
 } 
 return s;
}
int main()
{
 while(~scanf("%d",&n), n) 
 {
  int i ;
  memset ( c, 0, sizeof (c) );
  vector<int> v ;
  for(i = 0 ; i < n ; i++)
  {
   int t ;
   scanf("%d",&t);
   v.push_back(t); 
  }
  vector <int> p(v);
  sort ( p.begin (), p.end () );
  vector <int>::iterator end = unique ( p.begin(), p.end () );
  for ( i = 0; i < n; ++ i ) {
   v[i] = lower_bound ( p.begin(), end, v[i] ) - p.begin() + 1; 
  }
  long long res = 0;
  for ( i = n-1; i >= 0; -- i ) {
   res += sum ( v[i]-1 );
   mod ( v[i], 1 ) ;
  }
  printf ( "%lld\n", res );
 }
 return 0;
}