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 output0 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
5 9 1 0 5 4 3 1 2 3 0Sample Output
6 0
逆序数问题
采用归并排序的方法来将时间复杂度从O(n2)到O(nlogn)
要特别注意，由于数据非常大，因此最坏情况下答案是超出int范围的，要使用long long保存
AC代码：GitHub
1 /* 2 By:OhYee 3 Github:OhYee 4 HomePage:http://www.oyohyee.com 5 Email:oyohyee@oyohyee.com 6 Blog:http://www.cnblogs.com/ohyee/ 7 8 かしこいかわいい？ 9 エリーチカ！ 10 要写出来Хорошо的代码哦~ 11 */ 12 13 #include <cstdio> 14 #include <algorithm> 15 #include <cstring> 16 #include <cmath> 17 #include <string> 18 #include <iostream> 19 #include <vector> 20 #include <list> 21 #include <queue> 22 #include <stack> 23 #include <map> 24 using namespace std; 25 26 //DEBUG MODE 27 #define debug 0 28 29 //循环 30 #define REP(n) for(int o=0;o<n;o++) 31 32 const int maxn = 500005; 33 int a[maxn]; 34 35 long long ans; 36 37 //将已经排好序的a[l]~a[mid] a[mid+1]~a[r]拼合起来 38 void merge(int a[],int l,int mid,int r) { 39 int pos1 = l;//左侧的指针 40 int pos2 = mid + 1;//右侧的指针 41 int *temp = new int[r - l + 1]; 42 int pos = 0;//临时数组的指针 43 while(pos1 <= mid || pos2 <= r) { 44 if(pos1 > mid) { 45 temp[pos++] = a[pos2++]; 46 } 47 if(pos2 > r) { 48 temp[pos++] = a[pos1++]; 49 } 50 if(pos1 <= mid&&pos2 <= r) { 51 if(a[pos1] <= a[pos2]) { 52 temp[pos++] = a[pos1++]; 53 } else { 54 temp[pos++] = a[pos2++]; 55 ans += mid - pos1 + 1;//交换 56 } 57 } 58 } 59 for(int i = 0;i <= r - l;i++) 60 a[l + i] = temp[i]; 61 } 62 63 //归并排序 对a[l]~a[r]排序 64 void mergesort(int a[],int l,int r) { 65 if(l<r) { 66 int mid = (l + r) / 2; 67 mergesort(a,l,mid); 68 mergesort(a,mid + 1,r); 69 merge(a,l,mid,r); 70 } 71 } 72 73 bool Do() { 74 int n; 75 if(scanf("%d",&n),n == 0) 76 return false; 77 78 REP(n) 79 scanf("%d",&a[o]); 80 81 ans = 0; 82 mergesort(a,0,n - 1); 83 printf("%lld ",ans); 84 /*REP(n) 85 printf("%d ",a[o]); 86 printf(" ");*/ 87 88 return true; 89 } 90 91 int main() { 92 while(Do()); 93 return 0; 94 }