In this HackerEarth Special Shop problem solution, Creatnx now wants to decorate his house by flower pots. He plans to buy exactly N ones. He can only buy them from Triracle's shop. There are only two kind of flower pots available in that shop. The shop is very strange. If you buy X flower pots of kind 1 then you must pay A x X^2 and B x Y^2 if you buy Y flower pots of kind 2. Please help Creatnx buys exactly N flower pots that minimizes money he pays.

## HackerEarth Special Shop problem solution.

`#include <bits/stdc++.h>using namespace std;void solve() {    int test; cin >> test;    assert(1 <= test && test <= 1e5);    while (test--) {        int n, a, b; cin >> n >> a >> b;        assert(1 <= n && n <= 1e5);        assert(1 <= a && a <= 1e5);        assert(1 <= b && b <= 1e5);        int x = (int) round((long double) b * n / (a + b));        int y = n - x;        cout << (long long) a * x * x + (long long) b * y * y << "\n";    }}int main(int argc, char* argv[]) {    ios_base::sync_with_stdio(0), cin.tie(0);    if (argc > 1) {        assert(freopen(argv, "r", stdin));    }    if (argc > 2) {        assert(freopen(argv, "wb", stdout));    }    solve();    cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";    return 0;}`

### Second solution

`t = int(raw_input()) for ___ in xrange(t):    n,a,b = map(int, raw_input().split())     r1 = b * n / (a + b)     ans = 10101010101010101010    for x in xrange(r1-10,r1+10):        if 0 <= x <= n:            ans = min(ans, x*x*a + (n-x)*(n-x)*b)    print ans`