Merge sort-non-recursive application

/** function:     Merge 2 sequential arrays, effective to large     input:     Array First pdata[begin: mid]     Array Second pdata[mid+1..end]    output:pdata[begin...end]          Space required: O (n) **/void merge (int *pdata,int begin,int  mid,int end) {    int ptemp[8]={0};    int first  =begin;    int second =mid+1;    int itemp=begin //itemp start position cannot start at zero each time, Begin        while (first<=mid &&  second<=end)     {        if (PData[first] <=pdata[second])         {             pTemp[iTemp++]=pData[first++];        }else        {               pTemp[iTemp++]=pData[second++];         }    }        //Array to sort   No remaining      while (first<=mid && second>end)     {           pTemp[iTemp++]=pData[first++];    }         while (Second<=end && first>mid)      {       pTemp[iTemp++]=pData[second++];      }        for (int i=begin;i<=end;i++)     {         pdata[i]=ptemp[i];    }   / / delete&nbSp []ptemp;     }/***   function: Non-recursive form       Statistics I back door position, than I small number, joined [0--i]  statistics finished, I do not care about the back door is some unnecessary     premise: I do not have the statistics completed do not disrupt    1       How to do interval step access   observing rules       according to step search divided into n groups       step=1    [0,1], [2,3] ,[4,5][6,7] begin          step=2;    [40,80] [20,60]                         begin 0 mid=1 end  3           [40,80] [20,60] ,[41,81] [21,61]    begin 4 mid=5   end 7     step=4;            Group after a group of   may be less than one step of data     2   How to sort a step  data              **/void merge_sort (int * pdata,int length) {    int  step=0;//step  avoid all sorts, in order to determine whether first I do not fight with everyone     int begin=0;//the next split merge array start position       int mid=0;//begin-mid,mid-end  split 2 arrays to merge     int  end =0;//pending Sort End position     //int temp[length]={0};//cannot request space for temp   because length does not know         //control Merge Sort Count    for (step =1;step<length;step*=2)     {          cout<< "step==" <<step< < ":" <<endl;             //controls a merge sort, How many pairs of numbers need to be merged       //<=length must be equal to   less than an array of calculations        for (begin=0;Gin+2*step) <=length;begin=end+1)       {              mid=begin+step-1;           end=mid+step;          if (end>length-1)// Insufficient number of second series           {               end=length-1;           }           //cout<< "Begin_mid__end:" <<begin<< "__" <<mid<< "__" <<end<<endl;            merge (pdata,begin,mid,end);            //showdata (pdata,length);      }       showdata (pdata,length); &nBsp;        } }void testgetmaxsum () {      int array[8]={0,5,-2,1,-8,7,6,-3};      showdata (array,8);    merge_sort (array,8);       //showdata (array,12);    }

Output Result:

• Title Description:

• In the array of two digits, if the previous number is greater than the number that follows, then these two numbers form an inverse pair. Enter an array to find the total number of reverse pairs in this array.

• Input:

• Each test case consists of two lines:

  The first line contains an integer n, which represents the number of elements in the array. where 1 <= n <= 10^5.

  The second row contains n integers, and each array is of type int.

• Output:

• For each test case, output an integer that represents the total number of reverse pairs in the array.

Sample input:

47 5 6 4

• 
  

Sample output:

• 
  

5

• 
  

"Problem-solving ideas"

1 simulation manual calculation, disadvantage sequence execution

7 3

5 1

6 1

Sum=3+1+1=5

Cons: N*n

2 How to link up above. For example, student scores compare Class A and Class B if Class A first (i) about Class B (i) First place

comparison, you do not need to compare it with other values in B.

Merge sort-non-recursive application