Ultra-QuickSort POJ - 2299 树状数组求逆序对

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 
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

简单的树状数组应该  应该是最基本的应用求逆序对
注意离散化

 #include <cstdio>
 #include <cstring>
 #include <queue>
 #include <cmath>
 #include <algorithm>
 #include <set>
 #include <iostream>
 #include <map>
 #include <stack>
 #include <string>
 #include <vector>
 #define pi acos(-1.0)
 #define eps 1e-6
 #define fi first
 #define se second
 #define lson l,m,rt<<1
 #define rson m+1,r,rt<<1|1
 #define bug         printf("******\n")
 #define mem(a,b)    memset(a,b,sizeof(a))
 #define fuck(x)     cout<<"["<<x<<"]"<<endl
 #define f(a)        a*a
 #define sf(n)       scanf("%d", &n)
 #define sff(a,b)    scanf("%d %d", &a, &b)
 #define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
 #define pf          printf
 #define FRE(i,a,b)  for(i = a; i <= b; i++)
 #define FREE(i,a,b) for(i = a; i >= b; i--)
 #define FRL(i,a,b)  for(i = a; i < b; i++)
 #define FRLL(i,a,b) for(i = a; i > b; i--)
 #define FIN freopen("DATA.txt","r",stdin)
 #define lowbit(x)   x&-x
 #pragma comment (linker,"/STACK:102400000,102400000")

 using namespace std;
 typedef long long LL ;
 const int maxn = 5e5 + ;
 int n, a[maxn], c[maxn], b[maxn], bit[maxn];
 void update(int x) {
     while(x <= n) {
         bit[x] += ;
         x += lowbit(x);
     }
 }
 int sum(int i) {
     int ret = ;
     while(i > ) {
         ret += bit[i];
         i -= lowbit(i);
     }
     return ret;
 }
 int main() {
     while(scanf("%d", &n), n) {
         for (int i =  ; i <= n ; i++) {
             scanf("%d", &a[i]);
             b[i] = a[i];
         }
         sort(b + , b +  + n);
         for (int i =  ; i <= n ; i++)
             c[i] = lower_bound(b + , b +  + n, a[i]) - b;
         mem(bit, );
         LL ans=;
         for (int i= ;i<=n ;i++) {
             update(c[i]);
             ans+=i-sum(c[i]);
         }
         printf("%lld\n",ans);
     }
     return ;
 }