• # The Number of Inversions

For a given sequence A={a0,a1,...an1}A={a0,a1,...an−1}, the number of pairs (i,j)(i,j) where ai>ajai>aj and i<ji<j, is called the number of inversions. The number of inversions is equal to the number of swaps of Bubble Sort defined in the following program:

bubbleSort(A)
cnt = 0 // the number of inversions
for i = 0 to A.length-1
for j = A.length-1 downto i+1
if A[j] < A[j-1]
swap(A[j], A[j-1])
cnt++

return cnt

For the given sequence AA, print the number of inversions of AA. Note that you should not use the above program, which brings Time Limit Exceeded.

## Input

In the first line, an integer nn, the number of elements in AA, is given. In the second line, the elements aiai (i=0,1,..n1i=0,1,..n−1) are given separated by space characters.

## output

Print the number of inversions in a line.

## Constraints

• 1n200,0001≤n≤200,000
• 0ai1090≤ai≤109
• aiai are all different

5
3 5 2 1 4

6

3
3 1 2

## Sample Output 2

2
题意就是求一组数的逆序数，大小等于冒泡排序交换的次数，但看数据范围，这题用冒泡肯定超时。所以用归并排序模拟冒泡即可。
#include<iostream>
#include<cstring>
#include<stack>
#include<cstdio>
#include<cmath>
using namespace std;
#define MAX 500000
#define INF 2e9
int L[MAX/2+2],R[MAX/2+2];
long long  cnt=0;
long long merge(int A[],int n,int left,int mid,int right)
{
long long cnt=0;
int n1=mid-left;
int n2=right-mid;
for(int i=0;i<n1;i++)
{
L[i]=A[left+i];
}
for(int i=0;i<n2;i++)
{
R[i]=A[mid+i];
}
L[n1]=INF;
R[n2]=INF;
int i=0,j=0;
for(int k=left;k<right;k++)//合并
{
if(L[i]<=R[j])
A[k]=L[i++];
else
{
A[k]=R[j++];
cnt=cnt+(n1-i);
}
}
return cnt;
}
long long  mergeSort(int A[],int n,int left,int right)
{
long long v1,v2,v3;
if(left+1<right)
{
int mid=(left+right)/2;
v1=mergeSort(A,n,left,mid);
v2=mergeSort(A,n,mid,right);
v3=merge(A,n,left,mid,right);
return (v1+v2+v3);
}
else
return 0;
}
int main()
{
int A[MAX],n;
cnt=0;
cin>>n;
for(int i=0;i<n;i++)
cin>>A[i];
cnt=mergeSort(A,n,0,n);
cout<<cnt<<endl;
return 0;
}
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• 原文地址：https://www.cnblogs.com/hh13579/p/10833392.html