Ultra-QuickSort(归并排序)

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 49267		Accepted: 18035

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
题解，错了半天，while出错，改成for循环就对了，至今不知道为啥。。。
代码：

 #include<stdio.h>
 const int MAXN=;
 int a[MAXN],b[MAXN];
 long long ans;
 void mergesort(int start,int mid,int end){
     int i=start,k=start,j=mid+;
     while(i<=mid&&j<=end){
         if(a[i]<=a[j])b[k++]=a[i++];
         else{
             ans+=j-k;
             b[k++]=a[j++];
         }
     }
         while(i<=mid)b[k++]=a[i++];
         while(j<=end)b[k++]=a[j++];
         //while(start<=end)a[start++]=b[start];
         for(int i=start;i<=end;i++)a[i]=b[i];
 }
 void ms(int l,int r){
     if(l<r){
         int mid=(l+r)/;
         ms(l,mid);
         ms(mid+,r);
         mergesort(l,mid,r);
     }
 }
 int main(){
     int n;
     while(scanf("%d",&n),n){
         ans=;
         for(int i=;i<=n;i++)scanf("%d",a+i);
         ms(,n);
         printf("%lld\n",ans);
     }
     return ;
 }