Ultra-QuickSort
| Time Limit: 7000MS | Memory Limit: 65536K | |
| Total Submissions: 53685 | Accepted: 19722 |
Description
题目太长可以不看。目的是给定数列求逆序对。
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 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
5 9 1 0 5 4 3 1 2 3 0
Sample Output
6 0
Source
归并排序求逆序对。
归并排序分分钟写完，但是longlong忘了用lld输出，又浪费了一阵子青春
1 #include<algorithm> 2 #include<cstdio> 3 #include<iostream> 4 #include<cstring> 5 #include<cmath> 6 using namespace std; 7 const int mxn=520000; 8 long long a[mxn],t[mxn]; 9 int n; 10 long long ans; 11 void msort(int l,int r){ 12 if(r-l>1) 13 { 14 int mid=l+(r-l)/2; 15 msort(l,mid); 16 msort(mid,r); 17 int p=l,q=mid,i=l;//指向起点 18 while(p<mid || q<r){//范围内有数就继续处理 19 if(q>=r || (p<mid && a[p]<=a[q])) 20 { 21 t[i++]=a[p++]; 22 } 23 else {t[i++]=a[q++];ans+=mid-p;}; 24 } 25 for(i=l;i<r;i++)a[i]=t[i];//用排序后的序列覆盖原数组对应部分 26 } 27 return; 28 } 29 int main(){ 30 while(scanf("%d",&n) && n){ 31 ans=0; 32 int i,j; 33 for(i=1;i<=n;i++)scanf("%lld",&a[i]); 34 msort(1,n+1); 35 printf("%lld ",ans); 36 } 37 return 0; 38 }