题意很简单，就是用归并排序求逆序数。本想直接水过，没想到还调试了那么长时间，，最郁闷的是还tle了几次。。。。。。。题目：
Ultra-QuickSort
| Time Limit: 7000MS | Memory Limit: 65536K | |
| Total Submissions: 24782 | Accepted: 8862 |
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 issorted in ascending order. For the input sequence
Ultra-QuickSort produces the output
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.
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
ac代码：
#include <iostream>
#include <cstdio>
#include <string.h>
using namespace std;
#define M 500010
int a[M];
int b[M];
long long count;
void mergesort(int begin,int mid,int end){
int num=begin;
int i=begin,j=mid+1;
while((i<=mid)&&(j<=end)){
if(a[i]<=a[j])
b[num++]=a[i++];
else{
b[num++]=a[j++];
count+=(mid-i+1);
}
}
if(i>mid){
for(int k=j;k<=end;++k)
b[num++]=a[k];
}
else{
for(int k=i;k<=mid;++k){
b[num++]=a[k];
}
}
for(int k=begin;k<num;++k){
a[k]=b[k];
}
}
void merge(int begin,int end){
int mid;
if(begin<end){
mid=(begin+end)/2;
merge(begin,mid);
merge(mid+1,end);
mergesort(begin,mid,end);
}
}
int main(){
int n;
while(scanf("%d",&n),n){
count=0;
for(int i=1;i<=n;++i)
scanf("%d",&a[i]);
merge(1,n);
printf("%lld\n",count);
}
return 0;
}