POJ訓練計劃2299_Ultra-QuickSort(線段樹/單點更新)

POJ訓練計劃2299_Ultra-QuickSort(線段樹/單點更新)

解題報告

題意：

求逆序數。

思路：

線段樹離散化處理。

#include 
#include 
#include 
#include 
#define LL long long
using namespace std;
LL sum[2001000],num[501000],_hash[501000];
void push_up(int rt)
{
    sum[rt]=sum[rt*2]+sum[rt*2+1];
}
void update(int rt,int l,int r,int p,LL v)
{
    if(l==r)
    {
        sum[rt]=+v;
        return;
    }
    int mid=(l+r)/2;
    if(p<=mid)update(rt*2,l,mid,p,v);
    else update(rt*2+1,mid+1,r,p,v);
    push_up(rt);
}
LL q_sum(int rt,int l,int r,int ql,int qr)
{
    if(ql>r||qr


Ultra-QuickSort




Time Limit: 7000MS
 
Memory Limit: 65536K


Total Submissions: 41278
 
Accepted: 14952





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
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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