POJ 2299：Ultra-QuickSort

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

归并排序。

另外，此题有一坑就是结果会超int32；

详细能够參考：点击打开链接

我写的代码例如以下：

#include<cstdio>
#include<stdlib.h>
#include<cstring>
#include<algorithm>
#include<iostream>

using namespace std;

const int M = 500000 + 5;
int n, A[M], T[M], i;

long long merge_sort(int l, int r, int *A)
{
    if (r - l < 1) return 0;
    int mid = (l + r) / 2;
    long long ans = merge_sort(l, mid, A) + merge_sort(mid + 1, r, A);
    i = l;
    int p = l, q = mid + 1;
    while (p <= mid && q <= r)
    {
        if(A[p] <= A[q])
            T[i++] = A[p++];
        else
        {
            ans += (mid + 1 - p);
            T[i++] = A[q++];
        }
    }
    while (p <= mid) T[i++] = A[p++];
    while (q <= r) T[i++] = A[q++];
    for (int j = l; j <= r; j++)
        A[j] = T[j];
    return ans;
}

int main()
{
    int n;
    while(scanf("%d", &n) && n)
    {
        for(int j=0; j<n; j++)
            scanf("%d", &A[j]);
       printf("%lld\n", merge_sort(0, n - 1, A));
    }

    return 0;
}