Ultra-QuickSort
| Time Limit: 7000MS |
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Memory Limit: 65536K |
| Total Submissions: 46080 |
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Accepted: 16763 |
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
Source
Waterloo local 2005.02.05
#include
#include
#include
#include
#include
#include
using namespace std; int b[500005], c[500005]; int n; struct node { int num, id; } a[500005]; bool cmp(node a, node b) { return a.num < b.num; } int lowbit(int x) { return x&(-x); } void update(int i, int x) { while(i <= n) { c[i] += x; i = i + lowbit(i); } } int sum(int i) { int sum = 0; while(i > 0) { sum += c[i]; i = i - lowbit(i); } return sum; } int main() { int i; long long ans; while(scanf("%d", &n)!=EOF) { memset(b, 0, sizeof(b)); memset(c, 0, sizeof(c)); for(i = 1; i <= n; i++) { scanf("%d", &a[i].num); a[i].id = i; } sort(a+1, a+n+1, cmp); b[a[1].id] = 1; for(i = 2; i <= n; i++) { b[a[i].id] = i; } ans = 0; for(i = 1; i <= n; i++) { update(b[i], 1); ans += (sum(n)-sum(b[i])); } printf("%lld\n", ans); } return 0; }