Seeking to reverse the number (including the code and merge sorting algorithm) merge sort using

In the class, the teacher gives the algorithm design as a problem, so use merge sort algorithm just learning to achieve the required number of reverse.

So it is easy to think of the problem of violence based approach to O (n * n), but this is not what we need, therefore, to seek to achieve reverse merge sort of number, so we need to understand and master merge sort algorithm .

The so-called merge sort algorithm, is the use of divide and conquer the problem of scale continue to shrink, and then a process of consolidation.
Suppose there is an array, then we have to divide, until only a residual element, then back this time we merged, the process is very simple, is that each of the two array pointers motionless, specifically illustrated as follows:

then we implemented in code as follows:

#include<bits/stdc++.h>
#define maxn 10005
using namespace std;

int a[maxn];

//归并过程
void merge(int arr[], int l, int mid, int r){
	int help[r-l+1];//辅助数组
	int i = 0;
	int lIndex = l;
	int rIndex = mid+1;
	while(lIndex <= mid && rIndex <= r){
		help[i++] = arr[lIndex] < arr[rIndex] ? arr[lIndex++]:arr[rIndex++];	
	}
    //左边和右边肯定有一边到头了，不可能同时，因为每次只移动一边
	while(lIndex <= mid){
		help[i++] = arr[lIndex++];
	} 
	while(rIndex <= r){
		help[i++] = arr[rIndex++];
	}
    //将排好序的辅助数组赋值给原始数组，不需要返回值
	for(i = 0; i < r-l+1; i++){
		arr[l+i] = help[i];
	}
}
 
//递归
void mergeSort(int arr[], int l, int r){
	if(l == r){
		return;
	}
	int mid = (l + r) / 2;
    //左半部分归并排序
	mergeSort(arr, l, mid);
    //右半部分归并排序
	mergeSort(arr, mid+1, r);
    //左右部分归并
	merge(arr, l, mid, r);
}
 
//归并排序整个数组
void mergeSort(int arr[], int n){//特判，如果数组为空或只有一个元素，那么就不需要排序
	if(arr == NULL || n < 2){
		return;
	}
	mergeSort(arr,0,n-1);
}

int main(){
	int n; 
	while(cin >> n){
		for(int i = 0; i < n; i++) 
		cin >> a[i];
 
		mergeSort(a, n);
 
		for(int i = 0; i < n; i++){
			cout << a[i] << " ";
		} 
		cout << endl;
	}
	return 0;
} 

Then implement and master the basic merge sort algorithm, we just do it a little modification to meet the requirements of the subject.
Note: To enhance readability, let the teacher to run, I added a few lines of explanation;

/*
采用归并排序求逆序对数
测试数据:
5
1 2 3 4 5    0

5
5 4 3 2 1  10 
*/
#include<bits/stdc++.h>
#define maxn 10005
using namespace std;

int n;//输入数据的个数 
int tot;//用于计数求逆序对的个数 
int a[maxn];//用于存放数据个数 
int b[maxn];//存放 
 
void mersort(int l, int r){
	if(l>=r) return ;
	int midd = (l+r)/2;
	int i = l;
	int j = midd+1;
	int k = 0;
	while(i<=midd&&j<=r){
		if(a[i] > a[j]){
			b[k++] = a[j++];
			tot += midd-i+1; 
		}else{
			b[k++] = a[i++];
		}
	}
	while(i<=midd) b[k++] = a[i++];
	while(j<=r) b[k++] = a[j++];
	for(int i=0;i<k;i++) a[i+l] = b[i];
}

void mer(int l,int r){
	if(l>=r) return ;//超过则返回
	
	int mid = (l+r)/2; //不断划分 
	mer(l,mid);
	mer(mid+1,r);
	
	mersort(l,r);//合并 
}

int main(){
	cout<<"请输入数据的个数:"<<endl;
	cin>>n;
	cout<<"请输入数据,各数据间用空格隔开"<<endl;
	for(int i=0;i<n;i++)
	 cin>>a[i]; 
	tot = 0;//初始化为0 
	mer(0,n-1);
	cout<<"逆序对个数为:"<<endl; 
	cout<<tot<<endl; 
}