Ultra-QuickSort ————分治

n 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

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就是求逆序数对数

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#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
vector<int> A;
int n;
ll merge_count(vector<int> &a)
{
    int n=a.size();
    if(n<=1)    return 0;

    ll cnt=0;
    vector<int> b(a.begin(),a.begin()+n/2);
    vector<int> c(a.begin()+n/2,a.end());
    cnt += merge_count(b);
    cnt += merge_count(c);

    int ai=0,bi=0,ci=0;
    while(ai<n)
    {
        if(bi<b.size() && (ci == c.size() || b[bi] <= c[ci]))
            a[ai++] = b[bi++];
        else
        {
            cnt += n / 2 - bi;
            a[ai++] = c[ci++]; 
        }
    }
    return cnt;
}
int main()
{
    while(~scanf("%d",&n)&&n)
    {
        int x;
        A.clear();
        for(int i=0;i<n;i++)
        {
            scanf("%d",&x);
            A.push_back(x);
        }
        printf("%lld\n",merge_count(A));
    }
    return 0;
}