HackerEarth Plot the Curve problem solution

YASH PAL, 31 July 202414 February 2026

In this HackerEarth Plot, the Curve problem solution You are given with integers a,b,c,d,m. These represent the modular equation of a curve y*y mod m = (ax^3 + bx^2 + cx + d) mod m

Also, you are provided with an array A of size N. Now, your task is to find the number of pairs in the array that satisfy the given modular equation.

If (Ai,Aj) is a pair then Aj^2 mod m = (aAi^3 + bAi^2 + cAi + d) mod m.

Since the answer could be a very large output it modulo 10^9 + 7.

HackerEarth Plot the Curve problem solution

HackerEarth Plot the Curve problem solution.

#include <bits/stdc++.h>
#define M 1000000007
using namespace std;

int main() {
    
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int T;
    cin>>T;
    while(T--)
    {
        long long a,b,c,d,m;
        cin>>a>>b>>c>>d>>m;
        int N;
        cin>>N;
        int arr[N];
        unordered_map<int,long long> X,Y;
        for(int i=0;i<N;i++)
        {
            cin>>arr[i];
            long long x=((((((a*arr[i])%m)*arr[i])%m)*arr[i])%m+(((b*arr[i])%m)*arr[i])%m+(c*arr[i])%m+d%m)%m;
            long long y=(arr[i]*arr[i])%m;
            if(x<0)
                x=m+x;
            if(y<0)
                y=m+y;
            X[x]++;
            Y[y]++;
        }

        long long ans=0;
        for(auto it=X.begin();it!=X.end();it++)
            ans=(ans + (it->second * Y[it->first])%M)%M;
        cout<<ans<<endl;

    }
}

Second solution

#include <bits/stdc++.h>
 
using namespace std;
 
const int md = 1E9 + 7;
const int N = 1E5 + 5;
long long ar[N];
map<long long, int> mp;
 
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while(t --) {
        long long a, b, c, d, m;
        cin >> a >> b >> c >> d >> m;
        int n;
        cin >> n;
        for(int i = 0; i < n; i ++) {
            cin >> ar[i];
            mp[(((ar[i] * ar[i]) % m) + m) % m] ++;
        }
        long long ans = 0;
        for(int i = 0; i < n; i ++) {
            long long x = (((((((a * ar[i]) % m) * ((ar[i] * ar[i]) % m)) % m) + (b * ((ar[i] * ar[i]) % m) % m) + ((c * ar[i]) % m) + d) % m) + m) % m;
            if(mp.find(x) != mp.end())
                ans += mp[x];
        }
        cout << (ans % md) << 'n';
        mp.clear();
    } 
    return 0;
}

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