Codeforces Beta Round #69 (Div. 2 Only)

Codeforces Beta Round #69 (Div. 2 Only)

http://codeforces.com/contest/80

A

 #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 pb push_back
 #define eb emplace_back
 #define maxn 1000006
 #define eps 1e-8
 #define pi acos(-1.0)
 #define rep(k,i,j) for(int k=i;k<j;k++)
 typedef long long ll;
 typedef unsigned long long ull;

 int main(){
     #ifndef ONLINE_JUDGE
      //   freopen("input.txt","r",stdin);
     #endif
     std::ios::sync_with_stdio(false);
     int n,m;
     cin>>n>>m;
     for(int i=n+;;i++){
         int flag=;
         int j;
         for(j=;j<i;j++){
             if(i%j==){
                 break;
             }
         }
         if(j==i) {
             if(i==m) cout<<"YES"<<endl;
             else {
                 cout<<"NO"<<endl;
             }
             break;
         }
     }
 }

B

 #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 pb push_back
 #define eb emplace_back
 #define maxn 1000006
 #define eps 1e-8
 #define pi acos(-1.0)
 #define rep(k,i,j) for(int k=i;k<j;k++)
 typedef long long ll;
 typedef unsigned long long ull;

 int main(){
     #ifndef ONLINE_JUDGE
      //   freopen("input.txt","r",stdin);
     #endif
     std::ios::sync_with_stdio(false);
     double n,m;
     char ch;
     cin>>n>>ch>>m;
     if(n>=) n-=;
     double ans1,ans2;
     ans2=*m;
     ans1=n*+0.5*m;
     cout<<ans1<<" "<<ans2<<endl;
 }

C

因为有三个敌人，所以用三进制的方法枚举所有的情况。时间复杂度为O（3^7*7*7）

 #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 pb push_back
 #define eb emplace_back
 #define maxn 1000006
 #define eps 1e-8
 #define pi acos(-1.0)
 #define rep(k,i,j) for(int k=i;k<j;k++)
 typedef long long ll;
 typedef unsigned long long ull;

 map<string,ll>mp;
 string s[]={"Anka","Chapay","Cleo","Troll","Dracul","Snowy","Hexadecimal"};
 ll relation[][];

 int book[];

 int main(){
     #ifndef ONLINE_JUDGE
      //   freopen("input.txt","r",stdin);
     #endif
     std::ios::sync_with_stdio(false);
     int n;
     string s1,s2,s3;
     for(int i=;i<;i++){
         mp[s[i]]=i+;
     }
     cin>>n;
     for(int i=;i<=n;i++){
         cin>>s1>>s2>>s3;
         relation[mp[s1]][mp[s3]]++;
     }
     ll a,b,c;
     cin>>a>>b>>c;
     int m=******;
     set<int>se;
     ll ans1=0x3f3f3f3f;
     int ans2=;
     for(int i=;i<=m;i++){
         int j=i;
         se.clear();
         int co=;
         memset(book,,sizeof(book));
         while(j){
             book[co]=j%;
             se.insert(book[co]);
             co++;
             j/=;
         }
         if(se.size()==){
             int aa=,bb=,cc=;
             for(j=;j<=;j++){
                 if(book[j]==) aa++;
                 else if(book[j]==) bb++;
                 else if(book[j]==) cc++;
             }
             ll aaa=a/aa;
             ll bbb=b/bb;
             ll ccc=c/cc;
             if(aaa>bbb) swap(aaa,bbb);
             if(aaa>ccc) swap(aaa,ccc);
             if(bbb>ccc) swap(bbb,ccc);
             if(ccc-aaa<ans1){
                 ans1=ccc-aaa;
                 ans2=;
                 for(int q=;q<=;q++){
                     for(int w=;w<=;w++){
                         if(relation[q][w]&&book[q]==book[w]){
                             ans2++;
                         }
                     }
                 }
             }
             else if(ccc-aaa==ans1){
                 int tmp=;
                 for(int q=;q<=;q++){
                     for(int w=;w<=;w++){
                         if(relation[q][w]&&book[q]==book[w]){
                             tmp++;
                         }
                     }
                 }
                 ans2=max(ans2,tmp);
             }
         }
     }
     cout<<ans1<<" "<<ans2<<endl;
 }

D

要让delta>=0的情况为p-4q>=0。因此可以画出几何图形

以样例一为例子：红色部分就是需要求的概率

 #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 pb push_back
 #define eb emplace_back
 #define maxn 1000006
 #define eps 1e-8
 #define pi acos(-1.0)
 #define rep(k,i,j) for(int k=i;k<j;k++)
 typedef long long ll;
 typedef unsigned long long ull;

 int main(){
     #ifndef ONLINE_JUDGE
      //   freopen("input.txt","r",stdin);
     #endif
    // std::ios::sync_with_stdio(false);
     int t;
     scanf("%d",&t);
     while(t--){
         int a,b;
         scanf("%d %d",&a,&b);
         if(a==&&b==) printf("%.7f\n",*1.0);
         else if(a==&&b!=){
             printf("%.7f\n",0.5);
         }
         else if(a!=&&b==){
             printf("%.7f\n",*1.0);
         }
         else if(a>=*b){
             printf("%.7f\n",(a-b)*1.0/a);
         }
         else{
             printf("%.7f\n",(a/4.0+*b)*1.0/(*b));
         }
     }
 }

E

先做dfs，dfs的过程中对儿子可获得的最大果子进行排序，优先获取最大的果子，如果遍历完儿子之后自身的果子还有剩，就再对儿子进行一次遍历，这次遍历的目的是消耗儿子自身和自己的果实数量

 #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 pb push_back
 #define eb emplace_back
 #define maxn 1000006
 #define eps 1e-8
 #define pi acos(-1.0)
 #define rep(k,i,j) for(int k=i;k<j;k++)
 typedef long long ll;
 typedef unsigned long long ull;

 vector<int>ve[];
 int a[];
 int n;

 void dfs(int now,int pre,ll &Max,int &remain){
     vector<ll>ve_Max;
     vector<int>ve_remain;
     for(int i=;i<ve[now].size();i++){
         if(ve[now][i]!=pre&&a[ve[now][i]]>){
             ll tmp_Max;
             int tmp_remain=a[ve[now][i]]-;
             dfs(ve[now][i],now,tmp_Max,tmp_remain);
             ve_Max.pb(tmp_Max);
             ve_remain.pb(tmp_remain);
         }
     }
     sort(ve_Max.begin(),ve_Max.end());
     Max=;
     for(int i=ve_Max.size()-;i>=&&remain;i--){
         Max+=ve_Max[i]+;
         remain--;
     }
     for(int i=;i<ve_remain.size()&&remain;i++){
         Max+=min(remain,ve_remain[i])*;
         remain-=min(remain,ve_remain[i]);
     }
 }

 int main(){
     #ifndef ONLINE_JUDGE
      //   freopen("input.txt","r",stdin);
     #endif
     std::ios::sync_with_stdio(false);
     cin>>n;
     for(int i=;i<=n;i++) cin>>a[i];
     int u,v;
     for(int i=;i<n;i++){
         cin>>u>>v;
         ve[u].pb(v);
         ve[v].pb(u);
     }
     int fa;
     cin>>fa;
     ll ans=;
     dfs(fa,,ans,a[fa]);
     cout<<ans<<endl;
 }