#include #include #define rep(i,b) for(int i=0;i=0;i--) #define rep1(i,b) for(int i=1;i=x;i--) #define fore(i,a) for(auto& i:a) #define rng(x) (x).begin(), (x).end() #define rrng(x) (x).rbegin(), (x).rend() #define sz(x) ((int)(x).size()) #define pb push_back #define fi first #define se second #define pcnt __builtin_popcountll using namespace std; using namespace atcoder; using ll = long long; using ld = long double; template using mpq = priority_queue, greater>; template bool chmax(T &a, const T &b) { if (a bool chmin(T &a, const T &b) { if (b ll sumv(const vector&a){ll res(0);for(auto&&x:a)res+=x;return res;} bool yn(bool a) { if(a) {cout << "Yes" << endl; return true;} else {cout << "No" << endl; return false;}} #define retval(x) {cout << #x << endl; return;} #define cout2(x,y) cout << x << " " << y << endl; #define coutp(p) cout << p.fi << " " << p.se << endl; #define out cout << ans << endl; #define outd cout << fixed << setprecision(20) << ans << endl; #define outm cout << ans.val() << endl; #define outv fore(yans , ans) cout << yans << "\n"; #define outdv fore(yans , ans) cout << yans.val() << "\n"; #define coutv(v) {fore(vy , v) {cout << vy << " ";} cout << endl;} #define coutv2(v) fore(vy , v) cout << vy << "\n"; #define coutvm(v) {fore(vy , v) {cout << vy.val() << " ";} cout << endl;} #define coutvm2(v) fore(vy , v) cout << vy.val() << "\n"; using pll = pair;using pil = pair;using pli = pair;using pii = pair;using pdd = pair; using vi = vector;using vd = vector;using vl = vector;using vs = vector;using vb = vector; using vpii = vector;using vpli = vector;using vpll = vector;using vpil = vector; using vvi = vector>;using vvl = vector>;using vvs = vector>;using vvb = vector>; using vvpii = vector>;using vvpli = vector>;using vvpll = vector;using vvpil = vector; using mint = modint998244353; //using mint = modint1000000007; //using mint = dynamic_modint<0>; using vm = vector; using vvm = vector>; vector dx={1,0,-1,0,1,1,-1,-1},dy={0,1,0,-1,1,-1,1,-1}; ll gcd(ll a, ll b) { return a?gcd(b%a,a):b;} ll lcm(ll a, ll b) { return a/gcd(a,b)*b;} #define yes {cout <<"Yes"<>n; vi v(3); rep(i,3) v[i] = i; vl a(3),b(3); rep(i,3) cin>>a[i]>>b[i]; ll ans = 0; do{ rep(i,(int)1e4+1000){ ll rem = n - a[v[0]]*i; if (rem<0) break; ll mx = rem/a[v[1]]; rrepx(j,max(0ll,mx-2501),mx+1){ ll k = (rem - a[v[1]]*j) / a[v[2]]; if (k < 0) continue; ll cost = b[v[0]]*i + b[v[1]]*j + b[v[2]]*k; ll sum = a[v[0]]*i + a[v[1]]*j + a[v[2]]*k; assert(sum<=n); chmax(ans,cost); } } }while(next_permutation(rng(v))); out; return; } int main(){ ios::sync_with_stdio(false); cin.tie(0); int t = 1; //cin>>t; rep(i,t){ solve(); } return 0; }