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
5
9
1
0
5
4
3
1
2
3
0
Sample Output
6
0
题解：
树状数组求逆序数，先离散化，不然1e9开不下，然后每次对一个数id[i]，将树状数组中这个位置+1,统计这个位置的前缀和，然后用i-sum(id[i])(差实际上就是id[i]位置后缀和)就是这个数id[i]导致的逆序对数量。
代码:
1 #include<cstdio>
2 #include<iostream>
3 #include<cstring>
4 #include<algorithm>
5 using namespace std;
6 #define INF 0x3f3f3f3f
7 #define M(a, b) memset(a, b, sizeof(a))
8 const int N = 5e5 + 10;
9 int c[N], id[N], n;
10 struct node {
11 int val, pos;
12 bool operator < (const node &rhs) const {
13 return val < rhs.val;
14 }
15 }a[N];
16
17 int lowbit(int x) {return x & -x;}
18
19 void add(int x, int d) {
20 while (x <= n) {
21 c[x] += d;
22 x += lowbit(x);
23 }
24 }
25
26 int sum(int x) {
27 long long ret = 0;
28 while (x) {
29 ret += c[x];
30 x -= lowbit(x);
31 }
32 return ret;
33 }
34
35 int main() {
36 while (scanf("%d", &n), n) {
37 M(c, 0);
38 for (int i = 1; i <= n; ++i)
39 scanf("%d", &a[i].val), a[i].pos = i;
40 sort(a+1, a+1+n);
41 for (int i = 1; i <= n; ++i) id[a[i].pos] = i;
42 long long ans = 0;
43 for (int i = 1; i <= n; ++i) {
44 add(id[i], 1);
45 ans += i-sum(id[i]);
46 }
47 printf("%lld\n", ans);
48 }
49
50 return 0;
51 }