标签：long   bit   scan   i++   return   c++   逆序   pre   win   

AcWing 788. 逆序对的数量

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=1e6+10;
int q[N],tmp[N];
LL merge_sort(int l,int r){
    if(l>=r) return 0;
    int mid=l+r >>1;
    LL res=merge_sort(l,mid)+merge_sort(mid+1,r);
    int k=0,i=l,j=mid+1;
    while(i<=mid&&j<=r)
        if(q[i]<=q[j]) tmp[k++]=q[i++];
        else{
            tmp[k++]=q[j++];
            res+=mid-i+1;
        }
    while(i<=mid) tmp[k++]=q[i++];
    while(j<=r) tmp[k++]=q[j++];

    for(int i=l,j=0;i<=r;i++,j++) q[i]=tmp[j];
    return res;
}
int main(){
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
        scanf("%d",&q[i]);    
    cout<<merge_sort(0,n-1)<<endl;    
    return 0;
}

AcWing 788. 逆序对的数量

标签：long   bit   scan   i++   return   c++   逆序   pre   win   

原文地址：https://www.cnblogs.com/wiseXu/p/13390395.html