• 題意:對於一個數\(x\),有函式\(f(x)\),如果它是偶數,則\(x/=2\),否則\(x-=1\),不斷重複這個過程,直到\(x-1\),我們記\(x\)到\(1\)的這個過程為\(path(x)\),它表示這個過程中所有\(x\)的值,現在給你一個數\(n\)和\(k\),要你找出最大的數\(x\),並且\(x\)在\(path[1,n]\)中出現的次數不小於\(k\).

• 題解:我們對於\(x\)可以選擇分奇偶來討論.

  1.假如\(x\)為奇數,那麼根據題意,我們知道\(x,2x,2x+1,4x,4x+1,4x+2,8x,8x+1,8x+2,8x+3,8x+4....\)這些數包含且只有這些數包含\(x\)

  .
  2.假如\(x\)為偶數,那麼\(x,x+1,2x,2x+1,2x+2,2x+3,4x,4x+1,4x+2,4x+3,4x+4,4x+5,4x+6,4x+7,...\)這些數包含且只有這些數包含\(x\).
  那麼我們就可以分奇偶數來二分求答案.

• 程式碼:

#include <bits/stdc++.h>
#define ll long long
#define fi first
#define se second
#define pb push_back
#define me memset
#define rep(a,b,c) for(int a=b;a<=c;++a)
#define per(a,b,c) for(int a=b;a>=c;--a)
const int N = 1e6 + 10;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
using namespace std;
typedef pair<int,int> PII;
typedef pair<ll,ll> PLL;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b) {return a/gcd(a,b)*b;}

ll n,k;

bool check(ll x){
	ll tmp=x;
	ll cnt=0;
	ll cur;
	if(x&1) cur=1;
	else cur=2;
	while(x<=n){
		cnt+=min(cur,n-x+1);
		x<<=1;
		cur<<=1;
	}
	if(cnt>=k) return true;
	return false;
}

int main() {
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

	cin>>n>>k;

	ll l=0,r=n;

	ll ans=1;

	//odd:
	while(l<r){
		ll mid=(l+r+1)>>1;
		ll x=mid*2+1;
		if(check(x)) l=mid;
		else r=mid-1;
	}

	ans=max(ans,2*l+1);

	l=0,r=n;
	
	//even
	while(l<r){
		ll mid=(l+r+1)>>1;
		ll x=mid*2;
		if(check(x)) l=mid;
		else r=mid-1;
	}

	ans=max(ans,2*l);
	
	cout<<ans<<'\n';

    return 0;
}