CF959E Mahmoud and Ehab and the xor-MST 思维
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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 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
4
Output
Copy
4
Note
In the first sample:
?
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),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 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;
}
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