学步园

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.

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.

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.

5
9
1
0
5
4
3
1
2
3
0

6
0

 

这就是题求逆序数的题：方法有多种，最简单的用归并排序

我用的是树状数组；题目明显直接建树是不行的，因此得将其数变小，因此，用来两次快排

有个陷阱，和数据会超，得用long long 我就是因为这被ＷＡ了几次

 

#include<cstdio>
#include<algorithm>
#include<cstring>
#define N 500004
using namespace std;
int M;
long long f[N];
int s[N];
class node
{
  public:
  int num;
  int p;
}root[N];
int lowbit(int n)
{
  return n&(-n);
}
void add(int s)
{
  while(s<=M)
  {
    f[s]++;
    s+=lowbit(s);
  }
}
long long query(int num)
{
  long long sum=0;
  while(num>0)
  {
    sum+=f[num];
    num-=lowbit(num);
  }
  return sum;
}
bool cmp(node a,node b)
{
  return a.num<b.num;
}
bool cmpp(node a,node b)
{
  return a.p<b.p;
}
int main()
{
  int a;
  while(scanf("%d",&M)&&M)
  {
    memset(f,0,sizeof(f));
    long long sum=0;
    for(int i=0;i<M;i++)
    {
      scanf("%d",&root[i].num);
      root[i].p=i;
    }
    stable_sort(root,root+M,cmp);
    for(int i=0;i<M;i++)
    root[i].num=i+1;
    stable_sort(root,root+M,cmpp);
    for(int i=0;i<M;i++)
    {
      add(root[i].num);
      sum+=query(M)-query(root[i].num);
    }
    printf("%lld\n",sum);
  }
}