#include using namespace std; using ll=long long; using Pa=pair; using Vll=vector; using VVll=vector; using Vb=vector; using VVb=vector>; using Vs=vector; using VVs =vector>; using Vc=vector; using VVc =vector>; const ll MOD=1000000007; const ll INF=(ll)1<<60;// 10^18 < 1<<60 = 1152921504606846976 // VVll G(N+1,Vll(0)); // vector event; event.emplace_back(a,b); // 以下の関数 oi:エラー出力 OI:正出力 OIV:1次元vector正出力 OIVV:2次元vector正出力 template bool chmax(t&a,u b){if(a bool chmin(t&a,u b){if(bvoid oi(Head&& head,Tail&&... tail){if(sizeof...(tail)!=0){cerr<void OI(Head&& head,Tail&&... tail){if(sizeof...(tail)!=0){cout< void OIV(T &a){if(a.empty()){cout<<"\n"; return;} for(ll i00=0;i00<(a).size();i00++){if(a[i00]==(ll)1<<60){cout<<"I";} else{cout< void OIVV(T &a){if(a.empty()){cout<<"\n"; return;} for(ll i00=0;i00<(a).size();i00++){for(ll j00=0;j00<(a[i00]).size();j00++){if(a[i00][j00]==(ll)1<<60){cout<<"I";} else{cout<>N>>X>>M; Vll A(N+1,0); FOR(i,1,N +1) cin>>A[i]; Vll B(N+1,0);// 必要な回数 FOR(i,1,N +1){ ll a=A[i]; ll cnt=0; while(a>=X){ a/=2; cnt+=1; } B[i]=cnt; } ll ans=0;// 全員倒すのに何回使うのか FOR(i,1,N +1){ //oi(i,B[i],i*B[i]); chmax(ans,i*B[i]);// 消費kと回数 } if(ans<=M) OI("Yes"); else OI("No"); return 0; }