Submission #1312145

#TimeUsernameProblemLanguageResultExecution timeMemory
1312145cubedSelf Study (JOI22_ho_t2)C++20
100 / 100
97 ms6064 KiB
#include <bits/stdc++.h> using namespace std; #define ll long long #define yes cout<<"YES\n" #define no cout<<"NO\n" #define endl '\n' #define f first #define s second #define pb(x) push_back(x) #define int long long const int MOD = 1e9+7; const int inf =1e9; const ll INF = 1e18; const ll INV2 = 500000004; class DSU { vector<int> parent, size; public: DSU(int n) { parent.resize(n+1); size.resize(n+1); for (int i=0; i<=n; i++) { parent[i]=i; size[i]=1; } } int find (int node) { if (node==parent[node]) return node; return parent[node]=find(parent[node]); } void unite (int u, int v) { int pv=find(v); int pu=find(u); if (pv==pu) return; if (size[pu]<size[pv]) swap(pu, pv); parent[pv]=pu; size[pu]+=size[pv]; } int getsize (int node) { return size[find(node)]; } }; ll getsqrt(ll x) { ll val = sqrtl(x) + 2; while (val * val > x) val--; return val; } bool safe (int i, int j, int n, int m) { if (i<0 || j<0 || i>=n || j>=m) return false; return true; } ll gcd(ll a, ll b) { while (b != 0) { ll temp=b; b=a%b; a=temp; } return a; } ll lcm(ll a, ll b) { return a / gcd(a, b)*b; } vector<int> dx={-1, 0, 1, 0}; vector<int> dy={0, 1, 0, -1}; // SOLUTION STARTS FROM HERE // bool check (int x, int n, int m, vector<int> &a, vector<int>&b) { int tn = 0; int ta = n*m; for (int i=0; i<n; i++) { int slots =0; int best = max(a[i], b[i]); if (x<=m*best) { slots = (x+best-1)/best; } else { int rem = x-m*best; slots = m+(rem+b[i]-1)/b[i]; } tn+=slots; if (tn>ta) return false; } return tn<=ta; } void solve() { int n, m; cin>>n>>m; vector<int> a(n); for (int i=0; i<n; i++) { cin>>a[i]; } vector<int> b(n); for (int i=0; i<n; i++) { cin>>b[i]; } int l=0, r=1e18; int ans=0; while (l<=r) { int mid = l+(r-l)/2; if (check(mid, n, m, a, b)) { ans=max(ans, mid); l=mid+1; } else { r=mid-1; } } cout<<ans<<endl; // m=1, only one weak /* for subject i, if he studies ith subject in class x times and by himself y times then his score will be: x[i]*a[i] + y[i]+b[i]; we need to maximise the minimum value of this */ } bool multi=false; int32_t main() { ios_base::sync_with_stdio(false); cin.tie(nullptr); //freopen("convention2.in", "r", stdin); //freopen("convention2.out", "w", stdout); int t=1; if (multi) cin>>t; while (t--) solve(); return 0; }
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