poj-2299-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

59105431230

Sample Output

60

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给你一个未排序的数列，每次可换相邻的两个数，问换成一个正常升序后，所需要的次数。第一反应就是求逆序列，看了看时间给了7000ms，感觉水过去够了，于是水了一发，果不其然的超时了.....后来发现500000，o（n*n），根本水不过去，后来搜了一搜， 发现用归并排序可以做。于是做了

这是归并排序

void mergearray(int a[], int first, int mid, int last, int temp[])  {      int i = first, j = mid + 1;      int m = mid,   n = last;      int k = 0;            while (i <= m && j <= n)      {          if (a[i] <= a[j])              temp[k++] = a[i++];          else              temp[k++] = a[j++];      }            while (i <= m)          temp[k++] = a[i++];            while (j <= n)          temp[k++] = a[j++];            for (i = 0; i < k; i++)          a[first + i] = temp[i];  }  void mergesort(int a[], int first, int last, int temp[])  {      if (first < last)      {          int mid = (first + last) / 2;          mergesort(a, first, mid, temp);    //左边有序          mergesort(a, mid + 1, last, temp); //右边有序          mergearray(a, first, mid, last, temp); //再将二个有序数列合并      }  }    bool MergeSort(int a[], int n)  {      int *p = new int[n];      if (p == NULL)          return false;      mergesort(a, 0, n - 1, p);      delete[] p;      return true;  }  

下面是我的代码

    #include <iostream>    #include <cstdio>    #include <cstring>    #include <algorithm>    using namespace std;    long long num[500005];    long long temp[500005];    int n;    long long ans;    void merge(int low, int mid, int high)    {        int i = low, j = mid + 1, k = low;        while (i <= mid && j <= high)        {            if (num[i] <= num[j])            {                temp[k++] = num[i++];            }            else            {                ans += j - k ;                temp[k++] = num[j++];            }        }        while (i <= mid) temp[k++] = num[i++];        while (j <= high) temp[k++] = num[j++];        for (i = low; i <= high; ++i)        {            num[i] = temp[i];        }    }    void mergeSort(int a, int b)    {        if (a < b)        {            int mid = (a + b) / 2;            mergeSort(a, mid);            mergeSort(mid + 1, b);            merge(a, mid, b);        }    }    int main()    {        while (scanf("%d", &n) != EOF && n > 0)        {            ans = 0;            for (int i = 0; i < n; ++i)            {                scanf("%lld", num + i);            }            mergeSort(0, n - 1);            printf("%lld\n", ans);        }        return 0;    }