（POJ

（POJ - 2299）Ultra-QuickSort

Time Limit: 7000MS Memory Limit: 65536K

Total Submissions: 63706 Accepted: 23759

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

题目大意：给出一串数字求冒泡排序的次数。

思路：可以用归并排序做，因为我刚看树状数组，而这个就是求逆序对的和，所以下面用树状数组的方法求解。并在代码上给出详细注释。

#include<cstdio>#include<algorithm>#include<cstring>using namespace std;typedef long long LL;const int maxn=500005;int c[maxn],n;//存树状数组，n表示点数，从1开始int b[maxn];//存离散化后的数组struct node{    int val,id;}a[maxn];//存原始数据bool cmp(node a,node b){    return a.val<b.val;} //树状数组的三个函数int lowbit(int x){    return x&(-x);}void update(int x,int d)//在x处加d{    for(;x<=n;x+=lowbit(x)) c[x]+=d;}int getsum(int x){    int sum=0;    for(;x>0;x-=lowbit(x)) sum+=c[x];    return sum;} int main(){    while(~scanf("%d",&n)&&n)    {        for(int i=1;i<=n;i++)         {            scanf("%d",&a[i].val);            a[i].id=i;        }        //离散化         sort(a+1,a+n+1,cmp);        for(int i=1;i<=n;i++) b[a[i].id]=i;        memset(c,0,sizeof(c));        LL ans=0;        for(int i=1;i<=n;i++)//用树状数组求逆序数        {            update(b[i],1);            ans+=i-getsum(b[i]);        }         printf("%lld\n",ans);    }    return 0;}