求逆序数算法

#include<iostream>
#include<cstdlib>
#include<climits>
using namespace std;


 int Inversion(int a[],int start,int mid,int end){
     
     int PairNum = 0;
     
     for(int i=start;i<=mid;++i){
         
         for(int j=mid+1;j<=end;++j){
         
             if(a[i]>a[j])
                 ++PairNum;
         } 
     }
 
     return PairNum;
 }
 int Inversion1(int a[],int low,int mid,int high){
      int sub1_len=mid-low+1;                      
     int sub2_len=high-mid;
     int PairNum = 0;                      
     int *L=NULL,*R=NULL;
     L = (int *)malloc(sizeof(int)*(sub1_len));
     R = (int *)malloc(sizeof(int)*(sub2_len)); 
     for(int i=0;i<=sub1_len-1;++i){                
         L[i]=a[low+i];                          
     }
     for(int j=0;j<=sub2_len-1;++j){
         R[j]=a[mid+1+j];                       
     } 
      int i= 0; 
     int j =0;
      int k= low;
      while(i<sub1_len&&j<sub2_len){                //注意这里的循环条件，不是k<=high!!! 
          if(L[i]<=R[j])
          a[k++] = L[i++];
        else {
          a[k++] = R[j++];
          PairNum += sub1_len-i;                //当R数组元素r较小时 L中剩的元素个数是L中大于r的元素的个数， 累加起来即为有序的俩个子数组间的逆序数
        }
           
        
      }
      while(i<sub1_len)
         a[k++]= L[i++];
      while(j<sub2_len)
         a[k++]= R[j++];
      return PairNum;
 }
 
 int InversionPair(int a[],int start, int end){
     
     int mid = 0;
     int p,q,r;
     if(start<end){
         
         mid = (start+end)/2; 
         return InversionPair(a,start,mid)+InversionPair(a,mid+1,end)+Inversion1(a,start,mid,end);;
     }
     else
         return 0;
 }
 
 

int main(){
 
      int num;
      int InversionPairNum; 
      while(1){    
          cin>>num;
          InversionPairNum=0; 
          int *array=NULL;
          array = (int *)malloc(sizeof(int)*(num));
          for(int i=0;i<num;i++)    cin>>array[i];
          InversionPairNum = InversionPair(array,0,num-1);
          
          cout<<InversionPairNum<<endl;     
      }
 
     return 0;
 }