[POJ 2299]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

Source

Waterloo local 2005.02.05

tips：题面右边有个马桶塞不知道想表达什么。

简述题意，求一个数列的逆序对数量。多组数据，每组数据第一行一个n，接下来n行是这个数列。

题解：
由于a[i]很大，所以我们就不能用树状数组了。我们就得用归并排序来统计逆序对数量。

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#define ll long long
#define LiangJiaJun main
#define eps 1e-9
using namespace std;
int a[500004],temp[500004],n;
ll ans=0;
int combi(int l,int m,int r){
    int i=l,j=m+1,k=l;
    while(i<=m&&j<=r){
        if(a[i]>a[j]){
            ans+=(m-i+1);
            temp[k++]=a[j++];
        }
        else temp[k++]=a[i++];
    }
    while(i<=m)temp[k++]=a[i++];
    while(j<=r)temp[k++]=a[j++];
    for(i=l;i<=r;i++)a[i]=temp[i];
    return 0;
}
int erfen(int l,int r){
    if(r<=l)return 0;
    int mid=(l+r)>>1;
    erfen(l,mid);
    erfen(mid+1,r);
    combi(l,mid,r);
}
int LiangJiaJun(){
    while(scanf("%d",&n)!=EOF){
        ans=0;
        if(n == 0)break;
        for(int i=1;i<=n;i++)scanf("%d",&a[i]);
        erfen(1,n);
        printf("%lld\n",ans);
    }
    return 0;
}