硬核二分——cf985D

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+)>=2.0*n/mid)
        return ;
    return ;
}
int judge2(ll mid){//mid 是答案
    ll tmp=(mid-h+)/;
    ll sum=(h+h+tmp-)*tmp+(h-)*h/;
    if((mid-h+)%)
        sum+=(h+tmp);
    if(sum>=n)return ;
    return ;
}

int main(){
    cin>>n>>h;
    if((double)(h+)>=2.0*n/h){//h*(h+1)/2>=n
        ll l=,r=h,ans,mid;
        while(l<=r){
            mid=l+r>>;
            if(judge1(mid))
                ans=mid,r=mid-;
            else l=mid+;
        }
        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>>;
            if(judge2(mid))
                ans=mid,r=mid-;
            else l=mid+;
        }
        cout<<ans<<'\n';
    }
}