Codeforces Beta Round #9 (Div. 2 Only)

Codeforces Beta Round #9 (Div. 2 Only)

http://codeforces.com/contest/9

A

gcd水题

 #include<bits/stdc++.h>
 using namespace std;
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define sqr(x) ((x)*(x))
 #define maxn 1000010
 typedef long long ll;
 /*#ifndef ONLINE_JUDGE
         freopen("1.txt","r",stdin);
 #endif */
 int gcd(int a,int b){
     if(b==) return a;
     return gcd(b,a%b);
 }

 int main(){
     #ifndef ONLINE_JUDGE
       //  freopen("1.txt","r",stdin);
     #endif
     int n,m;
     cin>>n>>m;
     n=max(n,m);
     int fz=-n+;
     int fm=;
     int d=gcd(fz,fm);
    // cout<<fz<<" "<<fm<<endl;
     cout<<fz/d<<"/"<<fm/d<<endl;
 }

B

模拟题

 #include<bits/stdc++.h>
 using namespace std;
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define sqr(x) ((x)*(x))
 #define maxn 1000010
 typedef long long ll;
 /*#ifndef ONLINE_JUDGE
         freopen("1.txt","r",stdin);
 #endif */
 struct Point{
     ll x,y;
 }a[];

 double dist[][];

 int main(){
     #ifndef ONLINE_JUDGE
         freopen("1.txt","r",stdin);
     #endif
     ll n,vb,vs;
     cin>>n>>vb>>vs;
     for(int i=;i<=n;i++){
         cin>>a[i].x;
         a[i].y=;
     }
     ll sx,sy;
     cin>>sx>>sy;
     for(int i=;i<=n;i++){
         dist[][i]=sqrt(sqr(a[].x-a[i].x)+sqr(a[].y-a[i].y));
     }
     for(int i=;i<=n;i++){
         dist[i][]=sqrt(sqr(a[i].x-sx)+sqr(a[i].y-sy));
     }
     double ans=1e18;
     ll pos=;
     for(int i=;i<=n;i++){
         if(a[i].x!=){
             double t1=dist[][i]/vb;
             double t2=dist[i][]/vs;
             if(ans>=t1+t2){
                 ans=t1+t2;
                 pos=i;
             }
         }
     }
     cout<<pos<<endl;
 }

C

dfs

 #include<bits/stdc++.h>
 using namespace std;
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define sqr(x) ((x)*(x))
 #define maxn 1000010
 typedef long long ll;
 /*#ifndef ONLINE_JUDGE
         freopen("1.txt","r",stdin);
 #endif */
 map<ll,int>mp;
 ll n;
 int ans;

 void dfs(int pos){
     if(pos>n) return;
     if(!mp[pos]){
         ans++;
         mp[pos]=;
     }
     else return;
     dfs(pos*);
     dfs(pos*+);
 }

 int main(){
     #ifndef ONLINE_JUDGE
        // freopen("1.txt","r",stdin);
     #endif
     cin>>n;
     dfs();
     cout<<ans<<endl;
 }

D

参考博客：http://www.cnblogs.com/qscqesze/p/5414271.html

DP

dp[i][j]表示当前用了i个节点，高度小于等于j的方案数

dp[i][j] = sigma(dp[k][j-1]*dp[i-k-1][j-1])

 #include<bits/stdc++.h>
 using namespace std;
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define sqr(x) ((x)*(x))
 #define maxn 1000010
 typedef long long ll;
 /*#ifndef ONLINE_JUDGE
         freopen("1.txt","r",stdin);
 #endif */

 long long dp[][];

 int main(){
     #ifndef ONLINE_JUDGE
        // freopen("1.txt","r",stdin);
     #endif
     int n,h;
     cin>>n>>h;
     for(int i=;i<=n;i++){
         dp[][i-]=;
         for(int j=;j<=n;j++){
             for(int k=;k<j;k++){
                 dp[j][i]+=dp[k][i-]*dp[j-k-][i-];
             }
         }
     }
     cout<<dp[n][n]-dp[n][h-]<<endl;
 }

E

题意：给出n个点，m条边，问是否能通过加一些边，使得n个点构成有且仅有n条边的单个环

直接构造就好

 #include<bits/stdc++.h>
 using namespace std;
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define sqr(x) ((x)*(x))
 #define maxn 1000010
 typedef long long ll;
 /*#ifndef ONLINE_JUDGE
         freopen("1.txt","r",stdin);
 #endif */
 int fa[];
 int d[];
 vector<pair<int,int> > ve,ans;
 int Find(int x){
     int r=x,y;
     while(x!=fa[x]){
         x=fa[x];
     }
     while(r!=x){
         y=fa[r];
         fa[r]=x;
         r=y;
     }
     return x;
 }
 void join(int x,int y)
 {
     int xx=Find(x);
     int yy=Find(y);
     if(xx!=yy) fa[xx]=yy;
 }
 int main(){
     int n,m;
     cin>>n>>m;
     int v,u;
     for(int i=;i<=n;i++) fa[i]=i;
     for(int i=;i<=m;i++){
         cin>>u>>v;
         ve.push_back(make_pair(u,v));
         join(u,v);
         d[v]++,d[u]++;
         if(d[u]>||d[v]>){
             cout<<"NO"<<endl;
             return ;
         }
     }
     for(int i=;i<=n;i++){
         for(int j=;j<i;j++){
             if(d[j]<=&&d[i]<=&&Find(i)!=Find(j))
             {
                 ans.push_back(make_pair(j,i));
                 join(i,j);
                 d[i]++,d[j]++;
             }
         }
     }
     for(int i=;i<=n;i++){
         if(d[i]!=){
             for(int j=;j<i;j++){
                 if(d[j]==){
                     ans.push_back(make_pair(j,i));
                     join(i,j);
                     d[i]++,d[j]++;
                 }
             }
         }
     }
     for(int i=;i<=n;i++)
         if(d[i]==)ans.push_back(make_pair(i,i));
     int p = Find();
     for(int i=;i<=n;i++)
         if(Find(i)!=p){
             cout<<"NO"<<endl;
             return ;
         }
     cout<<"YES"<<endl;
     cout<<ans.size()<<endl;
     for(int i=;i<ans.size();i++)
         cout<<ans[i].first<<" "<<ans[i].second<<endl;
 }