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
59 1 0 5 431 2 3 0
Sample Output
6 0 给你一个排序,让你求出这个排序中的逆序数.算是自己写的树状数组的第一道题吧,这个题首先用到离散化思想.因为每个数都是小于1e9,而n的范围不超过5e5,而且我们只关心这些数的大小关系,所以我们就把这些数全部映射到1~n,然后再按照排好序的位置插进去就行了这样子是不会影响最后算逆序数的.映射后的序列为reflect接下来求逆序数,这时要用到树状数组,树状数组的功能是用来求数组前缀和的,我们假想有一个长度为n的数组,它现在每一位初始化为0我们按照位置顺序将reflect中的每一个数插入到树状数组中,假设我们现在插入的数是第i个数,它的值为x那么我就把刚刚假想的数组的第x位置变为1,即代表x被插入到该序列那么每一次对于ans怎样+呢?我们可以定义sum(x) x及小于x的数中有多少个元素已经被插入这样的话sum(x)就是我们刚刚那个假想数组的前缀和了,我们用树状数组可以求对于插入的第i个数,我们的ans+=i-sum(x),因为x是第i个数插进来的,之前插进来小于等于x的数为sum(x)两者做差就是在x之前插入,而且值要大于x的数字的个数代码如下:
1 #include <cstdio>
2 #include <cstring>
3 #include <cmath>
4 #include <algorithm>
5 using namespace std;
6 const int maxn =550000;
7 int reflect[maxn];
8 int c[maxn];
9 int n;
10 int lowbit (int x)
11 {
12 return x&(-x);
13 }
14 struct Node
15 {
16 int val,pos;
17 };
18 Node node[maxn];
19 bool cmp (Node q1,Node q2)
20 {
21 return q1.val<q2.val;
22 }
23
24 void update (int x)
25 {
26 while (x<=n){
27 c[x]+=1;//将我们想象的数组x位置附为1,直接就加到c[x]上了
28 x+=lowbit(x);//找父节点
29 }
30 }
31 int getsum (int x)
32 {
33 int sum=0;
34 while (x>0){
35 sum+=c[x];
36 /*求和时可以看作将x用二进制表示
37 temp=x;
38 每次加上c[temp],再把temp最后1位置上的值改成0
39 最后变成0时就求完了
40 如:求 101001
41 temp=101001 -> sum+=c[101001]
42 -> temp=101000 -> sum+=c[101000]
43 -> temp=100000 -> sum+=c[100000]
44 -> temp=000000 求完了
45 */
46 x-=lowbit(x);
47 }
48 return sum;
49 }
50 int main()
51 {
52 //freopen("de.txt","r",stdin);
53 while (scanf("%d",&n)&&n){
54 for (int i=1;i<=n;++i){
55 scanf("%d",&node[i].val);
56 node[i].pos=i;
57 }
58 sort(node+1,node + n + 1, cmp);
59 for (int i=1;i<=n;++i) reflect[node[i].pos]=i;//离散化映射到1~n
60 for (int i=0;i<=n;++i) c[i]=0;//初始化树状数组
61 long long ans=0;
62 for (int i=1;i<=n;++i){
63 update(reflect[i]);//插入一个数,更新一下
64 ans+=i-getsum(reflect[i]);//ans加一下
65 }
66 printf("%lld\n",ans);
67 }
68 return 0;
69 }