Ultra-QuickSort poj 2299 归并排序

Ultra-QuickSort

Time Limit: 7000MS Memory Limit: 65536KTotal Submissions: 35993 Accepted: 12965

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

交换相邻的元素，求至少交换多少次可以使其成递增的排序，如果用冒泡的话肯定会超时的，只能用归并排序，求其逆序数，

逆序数也就是它的交换数，所以用归并排序就是了！！！！

#include<iostream>#include<cstring>#include<cstdio>using namespace std;int a[500000];int sw[500000];int n;long long num;void merge(int low,int mid,int high){     int i=low;     int j=mid+1;     int m=0;     while(i<=mid&&j<=high)     {         if(a[i]<=a[j])         {             sw[m++]=a[i];             i++;         }         else         {             sw[m++]=a[j];             j++;             num=num+mid-i+1;         }     }     while(i<=mid)     {         sw[m++]=a[i];         i++;     }      while(j<=mid)     {         sw[m++]=a[j];         j++;     }     for(i=0;i<m;i++)     {         a[low+i]=sw[i];     }}void mergesort(int low,int high){     if(low<high)     {         int mid=(low+high)/2;         mergesort(low,mid);         mergesort(mid+1,high);         merge(low,mid,high);     }}int main(){    int i,j;    int n;    while(scanf("%d",&n)!=EOF)    {     if(n==0)     break;     num=0;     for(i=0;i<n;i++)     scanf("%d",&a[i]);     mergesort(0,n-1);     cout<<num<<endl;    }    return 0;}