"Divide and conquer" in array A, returns the number of I<j, and A[i]>a[j] (I,J) pairs

Title: EPI

In the subject, divide the array into the left and right parts, calculate the number of invrted pair of the left and right two parts respectively, then calculate a[i] and a[j] on one side of the case, then add the three.

int Count_invert_core (vector<int> &arr,int begin,int end) {if (End-begin + 1 < 2) return 0;if (End-begin + 1 = = 2) {if (Arr[begin] > Arr[end]) return 1;elsereturn 0;} At least 3 number int mid = begin + (end-begin) >>1;int num1 = Count_invert_core (arr, begin, mid); int num2 = Count_invert_cor  E (arr, mid + 1, end), sort (arr.begin () + Begin, Arr.begin () + mid + 1), sort (arr.begin () + mid+1, Arr.begin () + end + 1); int Add = 0;//add represents a number in the first half, a number of invrted pair in the second half for the (int i1 = begin, I2 = mid + 1; i1 <= mid; i1++) {while (i2<=end && Arr[i1] > Arr[i2]) i2++;if (i2 <= end) Add + = i2-1-Mid-1 + 1;elseadd + = end-mid-1 + 1;} return NUM1 + num2 + add;} int Count_invert (vector<int> &arr) {return Count_invert_core (Arr,0,arr.size ()-1);}

"Divide and conquer" in array A, returns the number of I<j, and A[i]>a[j] (I,J) pairs