10810-Ultra-QuickSort (merge sort to calculate the number of reverse orders)

Problem B: Ultra-QuickSortIn this problem, you have to analyze a particle 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 sequence

9 1 0 5 4,

Ultra-QuickSort produces the output

0 1 4 5 9.

Your task is to determine how 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 integerN< 500,000-- The length of the input sequence. Each of the followingNLines contains a single integer0 ≤ a [I] ≤ 999,999,999.,I-Th input sequence element. Input is terminated by a sequence of lengthN = 0. This sequence must not be processed.

For every input sequence, your program prints a single line containing an integer numberOp, The minimum number of swap operations necessary to sort the given input sequence.

Sample Input

59105431230

Output for Sample Input

60

The question is probably how many times to exchange the order ..

Idea: Use Merge Sorting to find the number of reverse orders in the sorting process. Time complexity O (nlog)

Code:

#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;}