HackerEarth Monk’s Encounter with Polynomial problem solution

YASH PAL, 31 July 202415 February 2026

In this HackerEarth Monk’s Encounter with Polynomial problem solution Our monk, while taking a stroll in the park, stumped upon a polynomial ( A X2 + B X +C ) lying on the ground. The polynomial was dying! Being considerate, our monk tried to talk and revive the polynomial.

The polynomial said:

I have served my purpose, and shall not live anymore. Please fulfill my dying wish. Find me the least non-negative integer Xo, that shall make my value atleast K i.e., A Xo2 + B Xo + C >= K .

Help our Monk fulfill the polynomial’s dying wish!

HackerEarth Monk's Encounter with Polynomial problem solution

HackerEarth Monk’s Encounter with Polynomial problem solution.

#include <bits/stdc++.h>
using namespace std;
#define ll long long int
int main()
{
    int test;
    cin>>test;
    while(test--)
    {
        ll A,B,C,K;
        cin>>A>>B>>C>>K;
        ll lb=0, ub =100000;
        ll ans = -1;
        while(lb<=ub)
        {
            ll mid = (lb+ub)/2;
            ll val = A*(mid*mid) + B*mid + C;
            if(val >= K){
                ans = mid;
                ub = mid-1;
            }
            else
                lb = mid+1;
        }
        cout<<ans<<endl;
    }
    return 0;
}

Second solution

#include<bits/stdc++.h>

using namespace std;

long long a,b,c;

long long f(long long x)
{
    return a*x*x + b*x + c;
}

int main()
{
    int t;
    long long k;
    cin>>t;
    assert(1 <= t && t <= 100000);
    while(t--)
    {
        cin>>a>>b>>c;
        cin>>k;
        assert(1 <= a && a <= 1000000);
        assert(1 <= b && b <= 1000000);
        assert(1 <= c && c <= 1000000);
        assert(1 <= k && k <= (long long)1e10);
        int ans = 100000000;
        int l = 0 , h = -b/(2*a) , m;
        while(l <= h)
        {
            m = (l+h)/2;
            if(f(m) >= k)
            {
                ans = min(ans,m);
                l = m + 1;
            }
            else
            {
                h = m - 1;
            }
        }
        l = max(0LL,-b/(2*a)) , h = 10000000;
        while(l <= h)
        {
            m = (l+h)/2;
            if(f(m) >= k)
            {
                ans = min(ans,m);
                h = m - 1;
            }
            else
            {
                l = m + 1;
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}

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