POJ2299 Ultra-QuickSort

A - Ultra-QuickSort

Time Limit:7000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice POJ 2299

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

/*Author: 2486Memory: 3880 KBTime: 360 MSLanguage: G++Result: Accepted*///序列的逆序数即为交换次数，所以求出该序列的逆序数即可//采用归并排序的思路进行求解，将他们分分划分为两个部分求解#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn=500000+5;int A[maxn],T[maxn];int n;long long cnt;void merge_sort(int *AA,int x,int y,int *NN){    if(y-x>1)    {        int m=x+(y-x)/2;        int p=x,q=m,i=x;        merge_sort(AA,x,m,NN);        merge_sort(AA,m,y,NN);        while(p<m||q<y)        {            if(q>=y||(p<m&&A[p]<=A[q]))NN[i++]=A[p++];//如果其中有一个为空的话，要判断正确，            else NN[i++]=A[q++],cnt+=m-p;//加入到了NN数组内的数字一定是比没有加入里面的小，因为，NN里面的都是排好序了的。所以m-p即为所求        }        for(i=x;i<y;i++)A[i]=NN[i];//将值附过去就可以了    }}int main() {    while(~scanf("%d",&n),n) {        cnt=0;        for(int i=0; i<n; i++) {            scanf("%d",&A[i]);        }        merge_sort(A,0,n,T);        printf("%I64d\n",cnt);    }    return 0;}