POJ---2299-Ultra-QuickSort

Ultra-QuickSort

Time Limit: 7000MS Memory Limit: 65536KTotal Submissions: 48797 Accepted: 17839

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

59105431230

Sample Output

60

Source

归并排序：

#include <cstdio>#include <cstring>#include <iostream>using namespace std;const int maxn=510000;int n,a[maxn],t[maxn];long long ans;void sort(int l,int r){    if(l==r)return;    int mid=(l+r)/2;    sort(l,mid);    sort(mid+1,r);    int i=l,j=mid+1,now=0;    while(i<=mid && j<=r)    {        if(a[i]>a[j])        {            ans+=mid-i+1;            t[++now]=a[j++];        }        else        {            t[++now]=a[i++];        }    }    while(i<=mid)t[++now]=a[i++];    while(j<=r)t[++now]=a[j++];    now=0;    for(int k=l; k<=r; ++k)        a[k]=t[++now];}int main(){    cin>>n;    while(n)    {        for(int i=1; i<=n; ++i)            scanf("%d",&a[i]);        ans=0;        sort(1,n);        cout<<ans<<endl;        scanf("%d",&n);    }}