牛客练习赛61题解
A.打怪
题意:
题解:
AC代码
#include<bits/stdc++.h> using namespace std; #define fi first #define se second #define pb push_back #define mp make_pair #define all(x) (x).begin(),(x).end() #define endl '\n' typedef long long ll; typedef pair<int, int> pii; typedef pair<ll, ll> pll; //const int mod=1e9+7; const int mod=998244353; const double eps = 1e-10; const double pi=acos(-1.0); const int maxn=3e6+10; const ll inf=0x3f3f3f3f; const int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}}; int main() { ios::sync_with_stdio(false); cin.tie(0);cout.tie(0); //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); int t; cin>>t; while(t--){ int h,a,H,A; cin>>h>>a>>H>>A; if(a>=H)cout<<-1; else { int k=(H-1)/a; int p=A*k; cout<<(h-1)/p; } cout<<endl; } return 0; }
B.吃水果
题意:
题解:
AC代码
#include<bits/stdc++.h> using namespace std; #define fi first #define se second #define pb push_back #define mp make_pair #define all(x) (x).begin(),(x).end() #define endl '\n' typedef long long ll; typedef pair<int, int> pii; typedef pair<ll, ll> pll; //const int mod=1e9+7; const int mod=998244353; const double eps = 1e-10; const double pi=acos(-1.0); const int maxn=3e6+10; const ll inf=0x3f3f3f3f; const int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}}; int main() { ios::sync_with_stdio(false); cin.tie(0);cout.tie(0); //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); int t; cin>>t; while(t--){ int a,b; cin>>a>>b; int ans=0; if(a>b)swap(a,b); while(1){ if(a==b){cout<<ans+a<<endl;break;} if(a*2<=b)a*=2,ans++; if(a*2>b)a--,b--,ans++; } } return 0; }
C.四个选项
题意:
题解:
AC代码
#include<bits/stdc++.h> using namespace std; #define fi first #define se second #define pb push_back #define mp make_pair #define all(x) (x).begin(),(x).end() #define endl '\n' typedef long long ll; typedef pair<int, int> pii; typedef pair<ll, ll> pll; //const int mod=1e9+7; const int mod=998244353; const double eps = 1e-10; const double pi=acos(-1.0); const int maxn=3e6+10; const ll inf=0x3f3f3f3f; const int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}}; vector<int> v; int a,b,c,d,m,n; int fa[20],cnt[20],ans; int find(int x){ if(fa[x]==x)return x; return fa[x]=find(fa[x]); } void dfs(int i,int xa,int xb,int xc,int xd){ if(xa==a&&xb==b&&xc==c&&xd==d)ans++; if(i==n)return; if(xa+v[i]<=a)dfs(i+1,xa+v[i],xb,xc,xd); if(xb+v[i]<=b)dfs(i+1,xa,xb+v[i],xc,xd); if(xc+v[i]<=c)dfs(i+1,xa,xb,xc+v[i],xd); if(xd+v[i]<=d)dfs(i+1,xa,xb,xc,xd+v[i]); } int main() { ios::sync_with_stdio(false); cin.tie(0);cout.tie(0); //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); cin>>a>>b>>c>>d>>m; for(int i=1;i<=12;i++)fa[i]=i; for(int i=1;i<=m;i++){ int x,y; cin>>x>>y; int p=find(x),q=find(y); fa[p]=q; } for(int i=1;i<=12;i++)cnt[find(i)]++; for(int i=1;i<=12;i++) if(cnt[i])v.pb(cnt[i]); n=v.size(); dfs(0,0,0,0,0); cout<<ans; return 0; }
D.最短路变短了
题意:
题解:
AC代码
#include<bits/stdc++.h> using namespace std; #define fi first #define se second #define pb push_back #define mp make_pair #define all(x) (x).begin(),(x).end() #define endl '\n' typedef long long ll; typedef pair<int, int> pii; typedef pair<ll, ll> pll; //const int mod=1e9+7; const int mod=998244353; const double eps = 1e-10; const double pi=acos(-1.0); const int maxn=1e5+10; const ll inf=0x3f3f3f3f; const int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}}; struct edge{ ll u,v,w; }e[200010]; int n,m; vector<pll> g1[maxn],g2[maxn]; ll d1[maxn],d2[maxn]; void dij(ll dis[],vector<pll> g[],int s){ for(int i=1;i<=n;i++)dis[i]=inf*inf; dis[s]=0; priority_queue<pll,vector<pll>,greater<pll> > q; q.push(mp(0,s)); while(!q.empty()){ pll p=q.top(); q.pop(); int u=p.se; if(dis[u]<p.fi)continue; for(auto i:g[u]){ int v=i.fi; ll w=i.se; if(dis[v]>dis[u]+w){ dis[v]=dis[u]+w; q.push(mp(dis[v],v)); } } } } int main() { ios::sync_with_stdio(false); cin.tie(0);cout.tie(0); //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); cin>>n>>m; for(int i=1;i<=m;i++){ ll u,v,w; cin>>u>>v>>w; g1[u].pb(mp(v,w)); g2[v].pb(mp(u,w)); e[i]={u,v,w}; } dij(d1,g1,1);dij(d2,g2,n); int q;cin>>q; while(q--){ ll x; cin>>x; if(d1[e[x].v]+d2[e[x].u]+e[x].w<d1[n])cout<<"YES"<<endl; else cout<<"NO"<<endl; } return 0; }