Codeforces Beta Round #69 (Div. 2 Only)
http://codeforces.com/contest/80
A
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define pb push_back 7 #define eb emplace_back 8 #define maxn 1000006 9 #define eps 1e-8 10 #define pi acos(-1.0) 11 #define rep(k,i,j) for(int k=i;k<j;k++) 12 typedef long long ll; 13 typedef unsigned long long ull; 14 15 16 int main(){ 17 #ifndef ONLINE_JUDGE 18 // freopen("input.txt","r",stdin); 19 #endif 20 std::ios::sync_with_stdio(false); 21 int n,m; 22 cin>>n>>m; 23 for(int i=n+1;;i++){ 24 int flag=1; 25 int j; 26 for(j=2;j<i;j++){ 27 if(i%j==0){ 28 break; 29 } 30 } 31 if(j==i) { 32 if(i==m) cout<<"YES"<<endl; 33 else { 34 cout<<"NO"<<endl; 35 } 36 break; 37 } 38 } 39 }
B
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define pb push_back 7 #define eb emplace_back 8 #define maxn 1000006 9 #define eps 1e-8 10 #define pi acos(-1.0) 11 #define rep(k,i,j) for(int k=i;k<j;k++) 12 typedef long long ll; 13 typedef unsigned long long ull; 14 15 16 int main(){ 17 #ifndef ONLINE_JUDGE 18 // freopen("input.txt","r",stdin); 19 #endif 20 std::ios::sync_with_stdio(false); 21 double n,m; 22 char ch; 23 cin>>n>>ch>>m; 24 if(n>=12) n-=12; 25 double ans1,ans2; 26 ans2=6*m; 27 ans1=n*30+0.5*m; 28 cout<<ans1<<" "<<ans2<<endl; 29 }
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）
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define pb push_back 7 #define eb emplace_back 8 #define maxn 1000006 9 #define eps 1e-8 10 #define pi acos(-1.0) 11 #define rep(k,i,j) for(int k=i;k<j;k++) 12 typedef long long ll; 13 typedef unsigned long long ull; 14 15 map<string,ll>mp; 16 string s[]={"Anka","Chapay","Cleo","Troll","Dracul","Snowy","Hexadecimal"}; 17 ll relation[15][15]; 18 19 int book[15]; 20 21 int main(){ 22 #ifndef ONLINE_JUDGE 23 // freopen("input.txt","r",stdin); 24 #endif 25 std::ios::sync_with_stdio(false); 26 int n; 27 string s1,s2,s3; 28 for(int i=0;i<7;i++){ 29 mp[s[i]]=i+1; 30 } 31 cin>>n; 32 for(int i=1;i<=n;i++){ 33 cin>>s1>>s2>>s3; 34 relation[mp[s1]][mp[s3]]++; 35 } 36 ll a,b,c; 37 cin>>a>>b>>c; 38 int m=3*3*3*3*3*3*3; 39 set<int>se; 40 ll ans1=0x3f3f3f3f; 41 int ans2=0; 42 for(int i=1;i<=m;i++){ 43 int j=i; 44 se.clear(); 45 int co=1; 46 memset(book,0,sizeof(book)); 47 while(j){ 48 book[co]=j%3; 49 se.insert(book[co]); 50 co++; 51 j/=3; 52 } 53 if(se.size()==3){ 54 int aa=0,bb=0,cc=0; 55 for(j=1;j<=7;j++){ 56 if(book[j]==0) aa++; 57 else if(book[j]==1) bb++; 58 else if(book[j]==2) cc++; 59 } 60 ll aaa=a/aa; 61 ll bbb=b/bb; 62 ll ccc=c/cc; 63 if(aaa>bbb) swap(aaa,bbb); 64 if(aaa>ccc) swap(aaa,ccc); 65 if(bbb>ccc) swap(bbb,ccc); 66 if(ccc-aaa<ans1){ 67 ans1=ccc-aaa; 68 ans2=0; 69 for(int q=1;q<=7;q++){ 70 for(int w=1;w<=7;w++){ 71 if(relation[q][w]&&book[q]==book[w]){ 72 ans2++; 73 } 74 } 75 } 76 } 77 else if(ccc-aaa==ans1){ 78 int tmp=0; 79 for(int q=1;q<=7;q++){ 80 for(int w=1;w<=7;w++){ 81 if(relation[q][w]&&book[q]==book[w]){ 82 tmp++; 83 } 84 } 85 } 86 ans2=max(ans2,tmp); 87 } 88 } 89 } 90 cout<<ans1<<" "<<ans2<<endl; 91 }
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
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define pb push_back 7 #define eb emplace_back 8 #define maxn 1000006 9 #define eps 1e-8 10 #define pi acos(-1.0) 11 #define rep(k,i,j) for(int k=i;k<j;k++) 12 typedef long long ll; 13 typedef unsigned long long ull; 14 15 16 int main(){ 17 #ifndef ONLINE_JUDGE 18 // freopen("input.txt","r",stdin); 19 #endif 20 // std::ios::sync_with_stdio(false); 21 int t; 22 scanf("%d",&t); 23 while(t--){ 24 int a,b; 25 scanf("%d %d",&a,&b); 26 if(a==0&&b==0) printf("%.7f ",1*1.0); 27 else if(a==0&&b!=0){ 28 printf("%.7f ",0.5); 29 } 30 else if(a!=0&&b==0){ 31 printf("%.7f ",1*1.0); 32 } 33 else if(a>=4*b){ 34 printf("%.7f ",(a-b)*1.0/a); 35 } 36 else{ 37 printf("%.7f ",(a/4.0+2*b)*1.0/(4*b)); 38 } 39 } 40 }
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
1 #include<bits/stdc++.h> 2 using namespace std; 3 #define lson l,mid,rt<<1 4 #define rson mid+1,r,rt<<1|1 5 #define sqr(x) ((x)*(x)) 6 #define pb push_back 7 #define eb emplace_back 8 #define maxn 1000006 9 #define eps 1e-8 10 #define pi acos(-1.0) 11 #define rep(k,i,j) for(int k=i;k<j;k++) 12 typedef long long ll; 13 typedef unsigned long long ull; 14 15 vector<int>ve[100005]; 16 int a[100005]; 17 int n; 18 19 void dfs(int now,int pre,ll &Max,int &remain){ 20 vector<ll>ve_Max; 21 vector<int>ve_remain; 22 for(int i=0;i<ve[now].size();i++){ 23 if(ve[now][i]!=pre&&a[ve[now][i]]>0){ 24 ll tmp_Max; 25 int tmp_remain=a[ve[now][i]]-1; 26 dfs(ve[now][i],now,tmp_Max,tmp_remain); 27 ve_Max.pb(tmp_Max); 28 ve_remain.pb(tmp_remain); 29 } 30 } 31 sort(ve_Max.begin(),ve_Max.end()); 32 Max=0; 33 for(int i=ve_Max.size()-1;i>=0&&remain;i--){ 34 Max+=ve_Max[i]+2; 35 remain--; 36 } 37 for(int i=0;i<ve_remain.size()&&remain;i++){ 38 Max+=min(remain,ve_remain[i])*2; 39 remain-=min(remain,ve_remain[i]); 40 } 41 } 42 43 44 int main(){ 45 #ifndef ONLINE_JUDGE 46 // freopen("input.txt","r",stdin); 47 #endif 48 std::ios::sync_with_stdio(false); 49 cin>>n; 50 for(int i=1;i<=n;i++) cin>>a[i]; 51 int u,v; 52 for(int i=1;i<n;i++){ 53 cin>>u>>v; 54 ve[u].pb(v); 55 ve[v].pb(u); 56 } 57 int fa; 58 cin>>fa; 59 ll ans=0; 60 dfs(fa,0,ans,a[fa]); 61 cout<<ans<<endl; 62 }