CF959E Mahmoud and Ehab and the xor-MST 思维

最新推荐文章于 2019-06-06 15:13:28 发布

转载最新推荐文章于 2019-06-06 15:13:28 发布 · 67 阅读

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原文链接：http://www.cnblogs.com/zxyqzy/p/10017081.html

本文探讨了一个结合位运算与特殊图论的问题，即在由n个顶点构成的完全图中，每条边的权重为连接两点的编号进行异或操作的结果，求该图的最小生成树总权重。通过对前几项的手动计算，发现每位上的贡献等于当前n的一半，从而得出解决此问题的有效算法。

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 vertices numbered from 0 to n - 1. For all 0 ≤ u < v < n, vertex u and vertex v are connected with an undirected edge that has weight $\text{[math]}$ (where $\text{[math]}$ 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 ≤ 10¹²), the number of vertices in the graph.

Output

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

Example

Input

Copy

Output

Copy

Note

In the first sample: $\text{[math]}$ The weight of the minimum spanning tree is 1+2+1=4.

题意翻译

n个点的完全图标号(0-n-1),i和j连边权值为i^j,求MST的值

不妨先手算几项，可以发现每一位上的贡献为当前n 的一半；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 300005
#define inf 0x3f3f3f3f
#define INF 9999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)

inline ll rd() {
	ll x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
	if (!b) {
		x = 1; y = 0; return a;
	}
	ans = exgcd(b, a%b, x, y);
	ll t = x; x = y; y = t - a / b * y;
	return ans;
}
*/



ll qpow(ll a, ll b, ll c) {
	ll ans = 1;
	a = a % c;
	while (b) {
		if (b % 2)ans = ans * a%c;
		b /= 2; a = a * a%c;
	}
	return ans;
}



int main()
{
	//ios::sync_with_stdio(0);
	ll n; rdllt(n);
	ll ans = 0;
	ll tmp = 1;
	while (n>1) {
		ans += tmp * (n >> 1); tmp <<= 1; n -= (n >> 1);
	//	cout << n<<' '<<ans << endl;
	}
	cout << ans << endl;
    return 0;
}

转载于:https://www.cnblogs.com/zxyqzy/p/10017081.html