「CF959E」Mahmoud and Ehab and the xor-MST【OEIS】

E. Mahmoud and Ehab and the xor-MST

time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output

Ehab is interested in the bitwise-xor operation and the special graphs. Mahmoud gave him a problem that combines both. He has a complete graph consisting of $n$n vertices numbered from 0 to $n - 1$n − 1. For all $0 ≤ u &lt; v &lt; n$0 ≤ u < v < n, vertex $u$u and vertex $v$v are connected with an undirected edge that has weight (where is the bitwise-xor operation). Can you find the weight of the minimum spanning tree of that graph?

You can read about complete graphs in https://en.wikipedia.org/wiki/Complete_graph

You can read about the minimum spanning tree in https://en.wikipedia.org/wiki/Minimum_spanning_tree

The weight of the minimum spanning tree is the sum of the weights on the edges included in it.

Input

The only line contains an integer $n (2 ≤ n ≤ 10^{12})$n(2 ≤ n ≤ 1012), the number of vertices in the graph.

Output

The only line contains an integer $x$x, the weight of the graph’s minimum spanning tree.

Example

input

4

output

4

Note

In the first sample: The weight of the minimum spanning tree is $1+2+1=4$1+2+1=4.

• 题意：

  给一个节点数为n的完全图，节点标号$[0,n-1]$[0,n−1]，边权定义为$u\ xor\ v$u xor v，$u,v$u,v为节点编号，求最小生成树

• 解法

  • 上次西安邀请赛就让我学会了一个东西：不会的东西，难直接推的东西直接打表！先打再说$233$233

  • 这题打表代码也贴一下吧

  •   #include<bits/stdc++.h>
      using namespace std;
      const int maxn=1e5+10;
      
      struct node{
      	int a,b,c;
      	node(int d=0,int e=0,int f=0){
      		a=d;b=e;c=f;
      	}
      	friend bool operator<(const node &a,const node &b){
      		return a.c<b.c;
      	}
      	void print(){
      		printf("%d %d %d\n",a,b,c);
      	}
      }a[maxn];
      
      int n;
      
      struct dsu{
      	int fa[maxn],rank[maxn];
      	void init(int k){
      		for(int i=1;i<=k;i++) fa[i]=i,rank[i]=0;
      	}
      	int fin(int k){
      		return fa[k]==k?k:(fa[k]=fin(fa[k]));
      	}
      	
      	bool unite(int a,int b){
      		int x=fin(a),y=fin(b);
      		if(x==y) return false;
      		if(rank[x]<rank[y]) 
      		fa[x]=y;
      		else{
      			fa[y]=x;
      			if(rank[x]==rank[y]) rank[x]++;
      		}
      		return true;
      	}
      	
      	bool same(int a,int b){
      		return fin(a)==fin(b);
      	}
      }tree;
      
      
      int main()
      {
      	while(~scanf("%d",&n)){
      		int tot=0;tree.init(n);
      		for(int i=0;i<n;i++){
      			for(int j=i+1;j<n;j++){
      				a[++tot]=node(i,j,i^j);
      			}
      		}
      		sort(a+1,a+tot+1);
      		//for(int i=1;i<=tot;i++) a[i].print();
      		int ans=0,cnt=0;
      		for(int i=1;i<=tot&&cnt<n-1;i++){
      			if(tree.unite(a[i].a+1,a[i].b+1)){
      				cnt++;
      				ans+=a[i].c;
      			}
      		}
      		printf("%d\n",ans);
      	}
      }
    

  • 然后找规律，找不出来直接$OEIS$OEIS吧$233$233

    • 

• 附代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n;

ll dfs(ll cur)
{
	if(cur==1) return 1;
	if(cur%2LL) return 2*dfs(cur/2)+cur/2+1;
	else return 2*dfs(cur/2)+cur/2;
}

int main()
{
	scanf("%lld",&n);
	printf("%lld\n",dfs(n-1));
}