[归并排序][逆序数]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

题意：要求使序列变为递增序列的最小交换相邻元素的次数；
思路：归并排序求逆序数；
AC代码：

#include <iostream>
#include<cstdio>
#include<algorithm>
#define N 500010
typedef long long ll;
using namespace std;

ll a[N];
ll ans=0;
ll tmp[N];

void merge_(int l,int m,int r){
  int cnt=0;
  int i,j;
  for(i=l,j=m+1;i<=m&&j<=r;){
    if(a[i]<=a[j]) {tmp[++cnt]=a[i]; i++;}
    else{
        tmp[++cnt]=a[j]; j++;
        ans+=(m-i+1);
    }
  }
  while(i<=m) {tmp[++cnt]=a[i]; i++;}
  while(j<=r) {tmp[++cnt]=a[j]; j++;}
  for(i=l,j=1;i<=r&&j<=cnt;i++,j++) a[i]=tmp[j];
}

void merge_sort(int l,int r){
  if(l==r) return;
  if(l<r){
    int m=(l+r)>>1;
    merge_sort(l,m);
    merge_sort(m+1,r);
    merge_(l,m,r);
  }
}

int main()
{
   int n;
   while(scanf("%d",&n)!=EOF&&n){
      for(int i=1;i<=n;i++) cin>>a[i];
      ans=0;
      merge_sort(1,n);
      cout<<ans<<endl;
   }
    return 0;
}

配图不是很懂啊？