POJ2299 Ultra-QuickSort(树状数组)

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本文介绍了一个特定的排序算法Ultra-QuickSort，并提供了一种高效计算输入序列逆序数的方法。通过离散化处理和二进制索引树（BIT），实现了对大规模整数序列逆序数的有效计算。

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Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 49181		Accepted: 17995

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

Sample Output

6
0

给出n个数，问你逆序数有多少对。注意数据范围，要将数据离散话并且ans要用long long。

AC代码：

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
using namespace std;
const int MAXN = 500005;
int n, c[MAXN], num[MAXN];
struct node
{
	/* data */
	int v, flag;
}a[MAXN];
bool cmp(node a, node b)
{
	return a.v < b.v;
}
int lowbit(int x) 
{
	return x & (-x);
}
void updata(int x)
{
	for(int i = x; i <= n; i += lowbit(i))
		c[i] += 1;
}
int get_sum(int x)
{
	int sum = 0;
	for(int i = x; i > 0; i -= lowbit(i))
		sum += c[i];
	return sum;
}
int main(int argc, char const *argv[])
{
	while(scanf("%d", &n) != EOF && n) {
		memset(c, 0, sizeof(c));
		for(int i = 1; i <= n; ++i) {
			scanf("%d", &a[i].v);
			a[i].flag = i;
		}
		sort(a + 1, a + 1 + n, cmp);
		for(int i = 1; i <= n; ++i)
			num[a[i].flag] = i;
		long long ans = 0;
		for(int i = 1; i <= n; ++i) {
			updata(num[i]);
			ans += i - get_sum(num[i]);
		}
		printf("%lld\n", ans);
	}
	return 0;
}