Ultra-QuickSort——归并排序

Ultra-QuickSort

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

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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

题意：

给出一组数，让这组数从小到大排序，问交换的次数

因为数据比较大，用归并排序能够保证不超时

#include <stdio.h>#include <string.h>#include <stdlib.h>int a[500009],L[500009],R[500009];__int64 s;void Merge(int l,int r,int m){    int i,j,k;    int n1=m-l+1;    int lenl = 0,len2 = 0;    for(i = l;i <= m;i++)        L[lenl++] = a[i];     for(i = m + 1;i <= r;i++)        R[len2++] = a[i];        L[lenl] = 99999999;        R[len2] = 99999999;        for(i = 0,j = 0,k = l;k <= r;k++)//以两个为一单元进行归并排序        {            if(L[i] <= R[j])                a[k] = L[i],i++;                else                    {                        a[k] = R[j],j++;                        s+=n1-i;//只要发现前边的有一个比后边的大，则从前变得这一个起，到最后都要与那个小的进行交换，次数增加                    }        }}void merge_sort(int s,int t){    if(s < t)    {        int m = (s + t)/2;        merge_sort(s,m);//将原数组拆分成最小单元        merge_sort(m+1,t);        Merge(s,t,m);    }}int main(){    int n;    int i,j,k;    while(scanf("%d",&n),n)    {        s=0;        for(i = 0;i < n;i++)            scanf("%d",&a[i]);        merge_sort(0,n-1);        printf("%I64d\n",s);    }    return 0;}