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**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[1], "r", stdin));

}

if (argc > 2) {

assert(freopen(argv[2], "wb", stdout));

}

solve();

cerr << "nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "msn";

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