Problem B: Ultra-QuickSort
In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of ndistinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence9 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.
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.
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
Output for Sample Input
6 0
题意：大概就是求排好序需要交换几次。。
思路：利用归并排序在排序过程中求出逆序数。时间复杂度O(nlog)
代码：
#include#include const int N = 500005; int n, num[N], save[N], sn; void init() { for (int i = 0; i < n; i++) scanf("%d", &num[i]); } long long solve(int l, int r, int *num) { if (r - l < 1) return 0; int mid = (l + r) / 2; long long ans = solve(l, mid, num) + solve(mid + 1, r, num); sn = l; int ll = l, rr = mid + 1; while (ll <= mid && rr <= r) { if (num[ll] <= num[rr]) save[sn++] = num[ll++]; else { ans += (mid + 1 - ll); save[sn++] = num[rr++]; } } while (ll <= mid) save[sn++] = num[ll++]; while (rr <= r) save[sn++] = num[rr++]; for (int i = l; i <= r; i++) num[i] = save[i]; return ans; } int main() { while (~scanf("%d", &n) && n) { init(); printf("%lld\n", solve(0, n - 1, num)); } return 0; }