Merge sort-Calculate reverse order number

Merge sort-Calculate reverse order number

Merge sort uses the thought of divide and conquer, Time complexity O (N*LOGN) dynamic memory utilization can reduce the space overhead;

Merge sort can also be used to calculate reverse order number;

Number of reverse order: a pair of numbers opposite the position and size in the sequence;

Number of reverse order = number of adjacent two digits exchanged in bubble sort;

intA[maxn],n;Long LongAns//The reverse number is generally very large, with a long longvoidCompute_ans (int*a,intBeginintMidintend) {    intlen_l=mid-begin+1; intlen_r=end-mid; int*left=New int[len_l+1];//Space Saving    int*right=New int[len_r+1];  for(intI=0; i<len_l;i++) left[i]=a[begin+i];  for(intI=0; i<len_r;i++) right[i]=a[mid+1+i]; Left[len_l]=right[len_r]=inf;//precautions against cross-border    inti,j;i=j=0;  for(intk=begin;k<=end;k++){        if(Left[i]<=right[j]) a[k]=left[i++]; Else{A[k]=right[j++]; Ans+=len_l-i;//calculates the number of reverse order, if only sort, the row is deleted; Reverse order = bubble Sort Interchange operation number}} delete left; //Remember to release, if not released will cause the MLEdelete right;}voidMergeSort (int*a,intBeginintend) {    if(begin<end) {        intMid= (begin+end)/2;        MergeSort (A,begin,mid); MergeSort (A,mid+1, end);    Compute_ans (A,begin,mid,end); }}intMain () { while(cin>>n,n) {ans=0;  for(intI=0; i<n;i++) cin>>A[i]; MergeSort (A,0, N-1); cout<<ans<<Endl; }    return 0;}

Merge Sort-Calculates the number of reverse orders

Merge sort-Calculate reverse order number