硬核二分——cf985D

Posted 2022-04-03 zsben991126

分两种情况进行讨论，要注意判条件时会有爆ll

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

int judge1(ll mid)//mid*(mid+1)>=2*n
    if((double)(mid+1)>=2.0*n/mid)
        return 1;
    return 0;
 
int judge2(ll mid)//mid 是答案  
    ll tmp=(mid-h+1)/2;
    ll sum=(h+h+tmp-1)*tmp+(h-1)*h/2;
    if((mid-h+1)%2)
        sum+=(h+tmp);
    if(sum>=n)return 1;
    return 0;


int main()
    cin>>n>>h;
    if((double)(h+1)>=2.0*n/h)//h*(h+1)/2>=n
        ll l=1,r=h,ans,mid;
        while(l<=r)
            mid=l+r>>1;
            if(judge1(mid))
                ans=mid,r=mid-1;
            else l=mid+1;
        
        cout<<ans<<‘\n‘;
    
    else //h*(h+1)/2 < n
        ll l=h,r=2e9,ans,mid;
        while(l<=r)//找到最小的符合条件的mid 
            mid=l+r>>1;
            if(judge2(mid))
                ans=mid,r=mid-1;
            else l=mid+1;
        
        cout<<ans<<‘\n‘;