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=500010;
int a[MAXN],b[MAXN];
long long ans;
void mergesort(int start,int mid,int end){
    int i=start,k=start,j=mid+1;
    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)/2;
        ms(l,mid);
        ms(mid+1,r);
        mergesort(l,mid,r);
    }
}
int main(){
    int n;
    while(scanf("%d",&n),n){
        ans=0;
        for(int i=1;i<=n;i++)scanf("%d",a+i);
        ms(1,n);
        printf("%lld
",ans);
    }
    return 0;
}