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```