1019 Nod number in reverse order (merge sort)

Title Link: https://www.51nod.com/onlineJudge/questionCode.html#!problemId=1019

Test instructions: In one arrangement, if the front and back positions of a pair of numbers are opposite the size order, that is, the previous number is greater than the number of subsequent numbers, then they are called an inverse. The total number of reverse order in a permutation is called the inverse number of the permutation.

such as 2 4 3 1, 2 1,4 3,4 1,3 1 is reverse, the reverse number is 4. An integer sequence is given to find the inverse number of the sequence. The inverse of the number of the two, one is in the process of bubble sequencing (direct violence enumeration, which is our normal O (n^2) level of the method) also is in the merge sort process, the time complexity can be reduced to O (NLOGN) on the merger sort of reverse order number, I lift a chestnut, A heap is (4 6 7 8), and the other pile is (2 6). We first merge the second pile of 2 into the first heap to find the first larger than its number 4, and then the resulting reverse number is len1-i+1=4-1+1=4 6 of the reverse number is 4-3+1=2. That's what this means .... The idea of merging sort is simple, which is to separate, order, and then merge. Be aware that updates to the data are good.

1#include <iostream>2#include <algorithm>3 using namespacestd;4 5 Const intn=50000+Ten;6typedefLong LongLL;7 8 LL CNT;9 intNum[n],result[n];Ten  One voidMergeint*num,int*result,intStartintend) { A     intT=start,mid= (start+end)/2; -     intI=start,j=mid+1; -      while(i<=mid&&j<=end) { the         if(Num[i]<=num[j]) result[t++]=num[i++]; -         Else if(num[i]>Num[j]) { -result[t++]=num[j++]; -cnt+=mid-i+1; +         } -     } +      while(I<=mid) result[t++]=num[i++]; A      while(j<=end) result[t++]=num[j++]; at      for(inti=start;i<=end;i++)//Update -num[i]=Result[i]; - } -  - voidMerge_sort (int*num,int*result,intStartintend) { -     if(end-start==1){ in         if(num[start]>Num[end]) { -             inttmp=Num[start]; tonum[start]=Num[end]; +num[end]=tmp; -cnt++; the         } *         return ; $     }Panax Notoginseng     if(end-start==0)return ; -Merge_sort (Num,result,start, (End+start)/2); theMerge_sort (Num,result, (End+start)/2+1, end); + merge (num,result,start,end); A } the  + intMain () { -     intN; $Cin>>N; $      for(intI=1; i<=n;i++) cin>>Num[i]; -Merge_sort (Num,result,1, n); -cout<<cnt<<Endl; the     return 0; -}

1019 Nod number in reverse order (merge sort)