Ultra-QuickSort
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Ultra-QuickSort
Time Limit: 7000MS Memory Limit: 65536KTotal Submissions: 57708 Accepted: 21325
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.
Ultra-QuickSort produces the output
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
#include <iostream>using namespace std;long long cnt;void merge(int array[],int left,int mid,int right)//排序+统计{ int* temp=new int[right-left+1]; int i,j,p; for(i=left,j=mid+1,p=0;i<=mid&&j<=right;p++) { if(array[i]<=array[j]) temp[p]=array[i++]; else { temp[p]=array[j++]; cnt+=(mid-i+1); } } while(i<=mid) temp[p++]=array[i++]; while(j<=right) temp[p++]=array[j++]; for(i=left,p=0;i<=right;i++) array[i]=temp[p++]; delete temp;}void mergesort(int array[],int left,int right)//二分{ if(left!=right) { int mid=(left+right)/2; mergesort(array,left,mid); mergesort(array,mid+1,right); merge(array,left,mid,right); }}int main(){ int n,array[500005]; while(cin>>n) { if(n==0) break; for(int i=0;i<n;i++) cin>>array[i]; cnt=0; mergesort(array,0,n-1); cout<<cnt<<endl; } return 0;}
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