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
題目大意：優化的冒泡排序，從小到大排序，就是找到一次沒有交換時BREAK，問我們至少交換多少次冒泡排序可完成。
思路：因為每次交換的都是左邊的數大於右邊的數，所以，我們需要求出有多少逆數對即可。可以利用歸並排序求解。
LANGUAGE：C
CODE：
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#include<iostream>
#include<cstdlib>
#include<cstdio>
using namespace std;
int a[500005],c[500005];
long long cnt;
void mergesort(int left,int right)
{
int mid,i,j,tmp;
if(right>left+1)
{
mid=(left+right)/2;
mergesort(left,mid);
mergesort(mid,right);
tmp=left;
for(i=left,j=mid;i<mid&&j<right;)
{
if(a[i]>a[j])
{
c[tmp++]=a[j++];
cnt+=mid-i;
}
else c[tmp++]=a[i++];
}
if(j<right)for(;j<right;++j)c[tmp++]=a[j];
else for(;i<mid;++i)c[tmp++]=a[i];
for(i=left;i<right;++i) a[i]=c[i];
}
}
int main()
{
//freopen("in.txt","r",stdin);
int n;
while(cin>>n,n)
{
cnt=0;
for(int i=1;i<=n;i++)
cin>>a[i];
mergesort(1,n+1);
cout<<cnt<<endl;
}
return 0;
}
