逆序数(归并排序)

#include <bits/stdc++.h>
using namespace std;

const int Maxn = 50010;
int A[Maxn];
int tmp[Maxn];
int cnt = 0;

void msort( int* A,int* tmp,int bg,int ed ){
    if( ed - bg > 1 ){
        int mid = bg + (ed-bg)/2;
        int x = bg;
        int y = mid;
        msort( A,tmp,bg,mid );
        msort( A,tmp,mid,ed );
        int index = bg;
        while( x < mid && y < ed ){
            if( A[x] > A[y] ){
                cnt += mid - x;//如果满足x < y and A[x] > A[y],即逆序,就加(mid-x),因为当前的A[x] > A[y],那么A[x]后面的所有数字都 > A[y],
                tmp[index++] = A[y++];
            }else{
                tmp[index++] = A[x++];
            }
        }
        while( x < mid ){
            tmp[index++] = A[x++];
        }
        while( y < ed ){
            tmp[index++] = A[y++];
        }
        for( int i = bg; i < ed; i++ ){
            A[i] = tmp[i];
        }
    }
}

int main(){
    int n;
    cin >> n;
    for( int i = 0; i < n; i++ ){
        cin >> A[i];
    }
    msort(A,tmp,0,n);
    cout << cnt << "\n";
}

def sortandcount( alist ):
    n = len( alist )
    count = 0
    if n == 2 or n == 1:
        if n == 2:
            if alist[0] > alist[1]:
                alist = alist[::-1]
                count += 1
        return alist,count
    a,cnt1 = sortandcount( alist[:n/2] )
    b,cnt2 = sortandcount( alist[n/2:] )
    count += cnt1 + cnt2
    resList = [0]*n
    i = j = 0
    len1 = len( a )
    len2 = len( b )
    for k in range( n ):
        if i < len1 and j < len2:
            if a[i] < b[j]:
                resList[k] = a[i]
                i += 1
            else:
                resList[k] = b[j]
                count += len1 - i;
                j += 1
        elif i < len1:
            resList[k] = a[i]
            i += 1
        else:
            resList[k] = b[j]
            j += 1
    return resList ,count

def main():
    n = input()
    num = [ input() for i in range(n) ]
    i,c = sortandcount(num)
    print c

if __name__ == ‘__main__‘:
    main()