Common Number

E - Common Number

Reference: codeforces Round # 608 (. Div's 2) E - the Common Number The (binary tree structure thought)

Specific practices may see reference blog.

After the key is divided parity, is the orderly

In a large range of data when you are trying to reduce the complexity of the answers dichotomy is a way to reduce the complexity, the complexity is reduced to O level (log (n)), so that even if the data is also 1e18 able to finish.

Code:

// Created by CAD on 2020/1/2.
#include <bits/stdc++.h>
#define fi first
#define se second
#define pll pair<long long,long long>
#define ll long long
using namespace std;

ll n,k;
bool judge(ll x)
{
    queue<pll> q;
    ll cnt=0;
    if(x&1) q.push({x,x});
    else q.push({x,x+1});
    while(q.size())
    {
        auto i=q.front();   q.pop();
        cnt+=min(n,i.se)-i.fi+1;
        if((i.fi<<1)<=n) q.push({i.fi<<1,i.se<<1|1});
    }
    return cnt>=k;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n>>k;
    ll ans=-1;
    //遍历偶数
    ll l=1,r=n/2;
    while(l<=r)
    {
        ll mid=(l+r)>>1;
        if(judge(mid<<1)) ans=max(ans,mid<<1),l=mid+1;
        else r=mid-1;
    }
    //遍历奇数
    l=0,r=n/2;
    while(l<=r)
    {
        ll mid=(l+r)>>1;
        if(judge(mid<<1|1)) ans=max(ans,mid<<1|1),l=mid+1;
        else r=mid-1;
    }
    cout<<ans<<endl;
    return 0;
}