Given an array of integers, if an element with a large subscript value in the array is less than the value of an element whose subscript value is smaller than it, it is called an inverse

Problem

Find out the number of reverse order

Given an integer array, the value of an element with a large subscript value in the array is less than the value of an element whose subscript value is smaller than it, which is called an inverse order. i.e.: array a[]; For i < J and A[i] > A[j], this is called an inverse order. Given an array, it is required to write a function that calculates the number of all the reverse order in the array.

"Code"

#include <stdio.h> #include <stdlib.h> #include <string.h>int sumnum = 0;void merge (int *a, int *b, int be gin, int mid, int end) {int I, j, k = 0;i = Begin;j = mid + 1;while (I <= mid && J <= end) {if (A[i] < a[j ]) b[k++] = a[i++];else {b[k++] = a[j++];sumnum + mid-i + 1;//calculate reverse order to}}while (I <= mid) b[k++] = A[i++];while (J <= E nd) b[k++] = a[j++];for (i = 0; i < K; i + +) A[i + begin] = B[i];} void mergesort (int *a, int *b, int begin, int end)//merge sort, time complexity O (N*LGN). {int mid = (begin + End)/2;if (begin < End) {MergeSort (A, B, begin, mid), MergeSort (A, B, mid + 1, end); Merge (A, B, be Gin, mid, End);}} int countreverse (int *a, int len)//Traversal lookup reverse order, Time complexity O (n^2).
{int count = 0;int I, j;for (i = 0; i < len-1; i + +) for (j = i + 1; j < Len; j + +) if (A[i] > a[j]) count++;return Count;} int main (void) {int Count;int a[] = {7, 4, 5, 3, 8, 6, 9, ten, 11};int len = sizeof (a)/sizeof (int), int *b;b = (int *) Mallo C (len * sizeof (int)); count = Countreverse (A, Len); MergeSort (A, B, 0, len-1);p rintf ("%d\n", Sumnum);p rintf ("%d\n", Count) ; free (b); return 0;}

"Execution Results"

Given an array of integers, if an element with a large subscript value in the array is less than the value of an element whose subscript value is smaller than it, it is called an inverse