HackerEarth Three rectangles problem solution

In this HackerEarth Three rectangles problem solution, You are given a rectangle of height H and width W. You must divide this rectangle exactly into three pieces such that each piece is a rectangle of integral height and width. You are required to minimize Area(max) – Area(min) where Area(max) is the area of the largest rectangle and Area(min) is the area of the smallest rectangle, among all three rectangle pieces.

HackerEarth Three rectangles problem solution.

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

#define     F           first
#define     S           second
#define     pb          push_back
#define     lb          lower_bound
#define     ub          upper_bound
#define     vi          vector<int>
#define     all(x)      x.begin(),x.end()
#define     fix         fixed<<setprecision(10)
#define     rep(i,a,b)  for(int i=int(a);i<=int(b);i++)
#define     repb(i,b,a) for(int i=int(b);i>=int(a);i--)
#define     FastIO      ios_base::sync_with_stdio(0),cin.tie(0)

typedef double db;
typedef long long ll;

const int N=2e5+5;
const int mod=1e9+7;

void solve(){
    ll h,w;
    cin>>h>>w;
    ll ans=1e18;
    rep(i,1,h-1){
        ll nh=h-i,nw=w;
        ll maxi=max({i*nw,nh/2*nw,(nh-nh/2)*nw});
        ll mini=min({i*nw,nh/2*nw,(nh-nh/2)*nw});
        ans=min(ans,maxi-mini);
        maxi=max({i*nw,nw/2*nh,(nw-nw/2)*nh});
        mini=min({i*nw,nw/2*nh,(nw-nw/2)*nh});
        ans=min(ans,maxi-mini);
    }
    rep(i,1,w-1){
        ll nw=w-i,nh=h;
        ll maxi=max({i*nh,nw/2*nh,(nw-nw/2)*nh});
        ll mini=min({i*nh,nw/2*nh,(nw-nw/2)*nh});
        ans=min(ans,maxi-mini);
        maxi=max({i*nh,nh/2*nw,(nh-nh/2)*nw});
        mini=min({i*nh,nh/2*nw,(nh-nh/2)*nw});
        ans=min(ans,maxi-mini);
    }
    cout<<ans<<'n';
}

signed main(){
    FastIO;
    int t;
    cin>>t;
    while(t--) solve();
    return 0;
}

Second solution

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

const int N = 1e5 + 14;

int solve(int h, int w) {
    int ans = INT_MAX;
    for (auto x : {(ll) w * h / (3 * h / 2), (ll) w * h / (3 * h / 2) + 1})
        ans = min<ll>(ans, max({abs(x * ll(h / 2) - ll(w - x) * h), x * ll((h + 1) / 2) - ll(w - x) * h, h % 2 * x}));
    return ans;
}

int main() {
    ios::sync_with_stdio(0), cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
        int h, w;
        cin >> h >> w;
        cout << min({h % 3 == 0 || w % 3 == 0 ? 0 : min(h, w), solve(h, w), solve(w, h)}) << 'n';
    }
}

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