Ultra-QuickSort (利用归并排序求逆序数)

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
可能会很不好想，其实这道题就是让求逆序数，原本我用的树状数组谁知道T了，用了归并排序才AC
AC code：

#include 
#include 
using namespace std;
const int MAXN = 500000+5;
long long cnt = 0;
int a[MAXN],p[MAXN];
void Merge_Array(int first, int mid, int last)
{
    int i=first, j=mid+1, k=first;
    while(i<=mid && j<=last)
    {
        if(a[i] <= a[j])
            p[k++] = a[i++];
        else
        {
            p[k++] = a[j++];
            cnt += j-k;
        }
    }
    while(i <= mid)
        p[k++] = a[i++];
    while(j <= last)
        p[k++] = a[j++];
    for(int i=first; i<=last; i++)
        a[i] = p[i];
}
void Merge_Sort(int first, int last)
{
    if(first < last)
    {
        int mid = (first + last) >> 1;
        Merge_Sort(first, mid);
        Merge_Sort(mid+1, last);
        Merge_Array(first, mid, last);
    }
}
int Scan()
{
    int res=0,ch,flag=0;
    if((ch=getchar())=='-')
        flag=1;
    else if(ch>='0'&&ch<='9')
        res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')
        res=res*10+ch-'0';
    return flag?-res:res;
}

void Out(long long a)
{
    if(a>9)
        Out(a/10);
    putchar(a%10+'0');
}
int main()
{
    int n;
    while(~scanf("%d",&n) && n)
    {
        cnt = 0;
        for(int i=0; i0, n-1);
        Out(cnt);
        puts("");
    }
    return 0;
}

TLE code：

#include 
#include 
#include 
#include 
#include 

using namespace std;

const int maxn = 5e5+5;

int bit[maxn];
int maxx;

int pos;

struct node {
    int val,id,v;
}ss[maxn];

int lowbit(int pos){
    return (pos & (-pos));
}

void update(int pos,int val) {
    while(pos <= maxx) {
        bit[pos] += val;
        pos += lowbit(pos);
    }
}

int query(int pos){
    int res = 0;
    while(pos > 0) {
        res += bit[pos];
        pos -= lowbit(pos);
    }
    return res;
}

int cmp(node x,node y) {
    return x.valint cmp1(node x,node y) {
    return x.idint main(){
    int t; 
    while(cin>>t && t)
    {
        map<int,int>mp;
        memset(bit,0,sizeof(bit));
        for (int i = 0;iscanf("%d",&ss[i].val);
            ss[i].id = i;
        }
        sort(ss,ss+t,cmp); pos = 0;
        for (int i = 0;iif ( !mp[ss[i].val] )
                mp[ss[i].val] = ++pos;
            ss[i].v = pos;
        } maxx = pos;
        sort(ss,ss+t,cmp1);
        int sum = 0;
        for (int i = 0;i1);
        }
        printf("%d\n",sum);
    }
    return 0;
}