Ultra-QuickSort POJ

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("******
")
#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 + 10;
int n, a[maxn], c[maxn], b[maxn], bit[maxn];
void update(int x) {
    while(x <= n) {
        bit[x] += 1;
        x += lowbit(x);
    }
}
int sum(int i) {
    int ret = 0;
    while(i > 0) {
        ret += bit[i];
        i -= lowbit(i);
    }
    return ret;
}
int main() {
    while(scanf("%d", &n), n) {
        for (int i = 1 ; i <= n ; i++) {
            scanf("%d", &a[i]);
            b[i] = a[i];
        }
        sort(b + 1, b + 1 + n);
        for (int i = 1 ; i <= n ; i++)
            c[i] = lower_bound(b + 1, b + 1 + n, a[i]) - b;
        mem(bit, 0);
        LL ans=0;
        for (int i=1 ;i<=n ;i++) {
            update(c[i]);
            ans+=i-sum(c[i]);
        }
        printf("%lld
",ans);
    }
    return 0;
}