poj2299：Ultra-QuickSort

求逆序对即可，然而发现还是bit容易写。。

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Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 49549		Accepted: 18142

Description

In this problem, you have to analyze a particular 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 many 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 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

5
9
1
0
5
4
3
1
2
3
0

Sample Output

6
0

Source

Waterloo local 2005.02.05

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define rep(i,n) for(int i=1;i<=n;i++)
struct data{
    ll w;int c;
    bool operator<(const data&rhs)const {
      return w<rhs.w;}
}e[500005];
int t[500005],n;
int lowbit(int x){return x&-x;}
void insert(int x){
    for(int i=x;i<=n;i+=lowbit(i))
      t[i]++;
}
int find(int x){
    int tmp=0;
    for(int i=x;i;i-=lowbit(i))
        tmp+=t[i];
    return tmp;
}
int main(){
    while(scanf("%d",&n)&&n){
        memset(t,0,sizeof(t));
        rep(i,n){
            scanf("%lld",&e[i].w);
            e[i].c=i;
        }
        sort(e+1,e+n+1);
        ll ans=0;
        rep(i,n){
            insert(e[i].c);
            ans+=i-find(e[i].c);
        }
        printf("%lld
",ans);
    }
    return 0;
}