Ultra-QuickSort

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

树状数组

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define lowbit(x) (x&(-x))
#define ls(x) (x<<1)//尽量宏定义吧，不太容易出错 
#define rs(x) (x<<1|1)
using namespace std;
typedef long long ll;
const int maxn=5e5+10;

struct node{
	int val,id;
	bool operator<(const node& b)const{
		return val<b.val;
	}
}val[maxn];
int aa[maxn];
ll tree[maxn];
int n;

void update(int val){
	while(val<=n){
		tree[val]+=1;
		val+=lowbit(val);
	}
}

ll query(int val){
	ll ans=0;
	while(val>0){
		ans+=tree[val];
		val-=lowbit(val);
	}
	return ans;
}

int main()
{
	while(scanf("%d",&n)&&n){
		for(int i=0;i<n;i++){
			scanf("%d",&val[i].val);
			val[i].id=i;
		}
		sort(val,val+n);
		
		ll cnt=1;
		aa[val[0].id]=cnt;
		for(int i=1;i<n;i++){
			if(val[i].val==val[i-1].val)
				aa[val[i].id]=cnt;
			else
				aa[val[i].id]=++cnt; 
		}
		memset(tree,0,sizeof(tree));
		ll ans=0;
		for(int i=0;i<n;i++){
			update(aa[i]);
			ans+=(i+1-query(aa[i]));
		}
		printf("%lld\n",ans);		 
	}	
	return 0;
}