poj2299 Ultra-QuickSort

D - Ultra-QuickSort

Time Limit:7000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 2299

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

原来是超快速排序，我说用快排怎么不对。他要的是冒泡排序的简单说就是逆序对，用归并就可以。

#include <iostream>#include<stdio.h>#include<stdlib.h>#include<string.h>#include<queue>#include<algorithm>using namespace std;int temp[500005];int a[500005];long long sum;void  hebing(int *a,int first,int med,int last){    int cur =0,i=first,j=med+1;    while(i<=med&&j<=last)    {        if(a[i]<=a[j])        {            temp[cur++]=a[i++];            continue;        }        else        {            sum+=med-i+1;            temp[cur++]=a[j++];        }    }    while(i<=med)temp[cur++]=a[i++];    while(j<=last)temp[cur++]=a[j++];    for(i=0; i<cur; i++)a[first++]=temp[i];}void gb(int *a,int first,int last){    if(first==last)return;    int med=(first+last)/2;    gb(a,first,med);    gb(a,med+1,last);    hebing(a,first,med,last);}int main(){    int n;    int i;    while(~scanf("%d",&n))    {        if(!n)break;        for(i = 0; i < n; ++i)scanf("%d",&a[i]);         sum=0;       gb(a,0,n-1);       printf("%lld\n",sum);    }    return 0;}