Reverse order pairs

Posted 2021-02-20 lym11248

In a sequence of numbers,we can find some number pairs that conform to the rule below:

Assuming that there exist two numbers,N and M and it‘s position in the sequence is n and m.

If the value of N is greater than M and n is less than m,we define the pair as "Reverse order pair".

So,try to find out all pairs in a sequence quickly.

 

Input:

5

5 2 1 3 4

 

Output:

5

 

S:

Of cause we can enumerate all the sequence.

But there is a better solution using the trait of merge sort.

Complete code is given below:

#include<iostream>
using namespace std;

int t[1000],a[1000];
int ans=0;
void mmerge(int ll, int rr) {
    if(ll==rr) return;
    else{
        int mid=(ll+rr)/2;
        mmerge(ll,mid);
        mmerge(mid+1,rr);
        int s=ll,q=ll,p=mid+1;
        while(s<=rr){
            if(q>mid||(p<=rr&&a[p]<a[q])){
                for(int m=q;m<=mid;m++)
                    cout<<"{"<<a[m]<<","<<a[p]<<"}"<<endl;
                t[s++]=a[p++];
　　　　　　　　　 ans+=mid-q+1;
　　　　　　　　　 //if the a[p] is less than a[q] it means that a[q]...a[mid] all greater than a[p].
            }
            else t[s++]=a[q++];
        }
        for(int i=ll;i<=rr;i++) a[i]=t[i];
    }
}
int main(){
    int N;
    cin>>N;
    for(int i=0;i<N;i++){
        cin>>a[i];
    }
    mmerge(0,N-1);
    cout<<endl<<ans;
    return 0;
}

Why does it work?

Because finding the pairs don‘t need the position of numbers.You don‘t have to maintain the position.

For example:

"52134" the reverse order pair of 1 are {5,1},{2,1},you can just make the judgment by the value of the number.Position is not cared.

"25134"----->{2,1},{5,1}.

So we can separate the sequence into groups.At first we count in small groups,then the larger groups.

In oder to compare the numbers,we sort the group in every layer.

Dichotomy+sort+merge=merge_sort.