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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> POJ 2299 Ultra-QuickSort £¨¹é²¢ÅÅÐò£©

POJ 2299 Ultra-QuickSort £¨¹é²¢ÅÅÐò£©

編輯:C++入門知識

POJ 2299 Ultra-QuickSort £¨¹é²¢ÅÅÐò£©


Ultra-QuickSort Time Limit: 7000MS Memory Limit: 65536K Total Submissions: 43446 Accepted: 15822

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 sZ†·Ÿ"http://www.Bkjia.com/kf/ware/vc/" target="_blank" class="keylink">vcnRlZCBpbiBhc2NlbmRpbmcgb3JkZXIuIEZvciB0aGUgaW5wdXQgc2VxdWVuY2UgPGJyPgo8Y2VudGVyPjkgMSAwIDUgNCAsPC9jZW50ZXI+Cjxicj4KVWx0cmEtUXVpY2tTb3J0IHByb2R1Y2VzIHRoZSBvdXRwdXQgPGJyPgo8Y2VudGVyPjAgMSA0IDUgOSAuPC9jZW50ZXI+Cjxicj4KWW91ciB0YXNrIGlzIHRvIGRldGVybWluZSBob3cgbWFueSBzd2FwIG9wZXJhdGlvbnMgVWx0cmEtUXVpY2tTb3J0IG5lZWRzIHRvIHBlcmZvcm0gaW4gb3JkZXIgdG8gc29ydCBhIGdpdmVuIGlucHV0IHNlcXVlbmNlLgo8cCBjbGFzcz0="pst">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 ¡Ü a[i] ¡Ü 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
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#include
#include
#define M 500010
int a[M];
int aux[M];
long long int  ans;
using namespace std;
void merge(int a[],int l,int mid,int h)
{
    for(int k=l;k<=h;++k)
        aux[k]=a[k];
    int i=l;
    int j=mid+1;
    for(int k=l;k<=h;++k)
    {
        if(i>mid)a[k]=aux[j++];
        else if(j>h)a[k]=aux[i++];
        else if(aux[i]=h)return ;
    int mid=l+(h-l)/2;
    sort(a,l,mid);
    sort(a,mid+1,h);
    merge(a,l,mid,h);
}
int main(int argc, char *argv[])
{
    //freopen("2299.in","r",stdin);
    int n;
    while(scanf("%d",&n)&&n!=0){
        ans=0;
        for(int i=0;i

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