Codeforces Round #514 (Div. 2)
https://codeforces.com/contest/1059
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 10000005 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 pair<int,int> pii; 14 typedef pair<long long,int>pli; 15 typedef pair<int,char> pic; 16 typedef pair<pair<int,string>,pii> ppp; 17 typedef unsigned long long ull; 18 const long long mod=1e9+7; 19 const double oula=0.57721566490153286060651209; 20 using namespace std; 21 22 int n,L,k; 23 vector<pii>ve; 24 25 int main(){ 26 std::ios::sync_with_stdio(false); 27 cin>>n>>L>>k; 28 int x,y; 29 for(int i=1;i<=n;i++){ 30 cin>>x>>y; 31 ve.pb({x,y}); 32 } 33 int ans=0; 34 int pos=0; 35 for(int i=0;i<ve.size();i++){ 36 if(ve[i].first-pos>=k) ans+=(ve[i].first-pos)/k; 37 pos=ve[i].first+ve[i].second; 38 } 39 if(L-pos>=k){ 40 ans+=(L-pos)/k; 41 } 42 cout<<ans<<endl; 43 44 45 }
B
模拟
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 10000005 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 pair<int,int> pii; 14 typedef pair<long long,int>pli; 15 typedef pair<int,char> pic; 16 typedef pair<pair<int,string>,pii> ppp; 17 typedef unsigned long long ull; 18 const long long mod=1e9+7; 19 const double oula=0.57721566490153286060651209; 20 using namespace std; 21 22 int n,m; 23 string str[1005]; 24 bool book[1005][1005]; 25 26 void Check(int x,int y){ 27 if(str[x][y]=='#'&&str[x][y+1]=='#'&&str[x][y+2]=='#'&&str[x+1][y]=='#'&&str[x+1][y+2]=='#'&&str[x+2][y]=='#'&&str[x+2][y+1]=='#'&&str[x+2][y+2]=='#'){ 28 book[x][y]=1;book[x][y+1]=1;book[x][y+2]=1;book[x+1][y]=1;book[x+1][y+2]=1;book[x+2][y]=1;book[x+2][y+1]=1;book[x+2][y+2]=1; 29 } 30 } 31 32 int main(){ 33 std::ios::sync_with_stdio(false); 34 cin>>n>>m; 35 for(int i=0;i<n;i++) cin>>str[i]; 36 for(int i=0;i<n-2;i++){ 37 for(int j=0;j<m-2;j++){ 38 Check(i,j); 39 } 40 } 41 for(int i=0;i<n;i++){ 42 for(int j=0;j<m;j++){ 43 if(str[i][j]=='#'){ 44 if(!book[i][j]){ 45 cout<<"NO"<<endl; 46 return 0; 47 } 48 } 49 } 50 } 51 cout<<"YES"<<endl; 52 }
C
数论
题意：给出n,有值为1-n的数字，每去掉一个值后输出剩下值的gcd，使输出的字典序最大
思路：因为相邻的两个数互质，所以要先把奇数都去掉，剩下的都是2的倍数，然后再去掉2,6,10..这样的数，使剩下的都是4的倍数，以此类推
有种情况是剩下的数为3个，比如 x,2x,3x,这样的话，gcd的值为x,x,3x
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 10000005 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 pair<int,int> pii; 14 typedef pair<long long,int>pli; 15 typedef pair<int,char> pic; 16 typedef pair<pair<int,string>,pii> ppp; 17 typedef unsigned long long ull; 18 const long long mod=1e9+7; 19 const double oula=0.57721566490153286060651209; 20 using namespace std; 21 22 23 int main(){ 24 std::ios::sync_with_stdio(false); 25 int n; 26 cin>>n; 27 int nn=(n+1)/2; 28 int ans=1; 29 if(n==3){ 30 cout<<ans<<" "<<ans<<" "<<ans*3<<endl; 31 return 0; 32 } 33 while(n){ 34 for(int i=1;i<=nn;i++) cout<<ans<<" "; 35 ans<<=1; 36 n>>=1; 37 nn=(n+1)/2; 38 if(n==3){ 39 cout<<ans<<" "<<ans<<" "<<ans*3<<endl; 40 break; 41 } 42 } 43 }
D
二分+几何
题意：给定n个点，求最小圆半径，使圆可以覆盖这n个点且与x轴相交
思路：二分圆的半径，遍历每个点，求该点可以被圆覆盖的区间，最后看看区间是否有交集
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 10000005 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 pair<int,int> pii; 14 typedef pair<double,double>pdd; 15 typedef pair<int,char> pic; 16 typedef pair<pair<int,string>,pii> ppp; 17 typedef unsigned long long ull; 18 const long long mod=1e9+7; 19 const double oula=0.57721566490153286060651209; 20 using namespace std; 21 22 vector<pdd>ve; 23 24 int sgn(double x){ 25 if(x<0) return -1; 26 if(x<eps) return 0; 27 return 1; 28 } 29 30 bool Check(double mid){ 31 double L=-100000000000000000.0,R=100000000000000000.0,dist; 32 for(int i=0;i<ve.size();i++){ 33 if(ve[i].second>mid*2) return false; 34 dist=sqrt(2*mid*ve[i].second-ve[i].second*ve[i].second); 35 if(L<ve[i].first-dist) L=ve[i].first-dist; 36 if(R>dist+ve[i].first) R=dist+ve[i].first; 37 } 38 return L<R; 39 } 40 41 int main(){ 42 double x,y; 43 int n; 44 scanf("%d",&n); 45 int flag1=0,flag2=0; 46 for(int i=1;i<=n;i++){ 47 scanf("%lf %lf",&x,&y); 48 ve.pb({x,y}); 49 if(y>0) flag1=1; 50 else if(y<0) flag2=1; 51 } 52 if(flag1&&flag2){ 53 puts("-1"); 54 return 0; 55 } 56 for(int i=0;i<ve.size();i++){ 57 ve[i].second=fabs(ve[i].second); 58 } 59 double L=0,R=100000000000000000.0,mid; 60 for(int i=1;i<=100;i++){///因为精度问题调不好，所以直接用for循环。。。 61 mid=(L+R)/2; 62 if(Check(mid)){ 63 R=mid; 64 } 65 else{ 66 L=mid; 67 } 68 } 69 printf("%.7f ",L); 70 71 }
E
贪心+倍增
题意：问一棵树最少可以被分为多少条垂直路径，路径长小于等于L，点权值之和小于等于S
思路：一个个点暴力过去会超时，所以用倍增法，看看一个点最多可以向上走多少个点，符合条件就选择
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 100005 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 pair<int,int> pii; 14 typedef pair<double,double>pdd; 15 typedef pair<int,char> pic; 16 typedef pair<pair<int,string>,pii> ppp; 17 typedef unsigned long long ull; 18 const long long mod=1e9+7; 19 const double oula=0.57721566490153286060651209; 20 using namespace std; 21 22 ll n,L,S; 23 ll a[100005]; 24 vector<int>ve[100005]; 25 ll parent[maxn][35],val[maxn][35]; 26 int len[maxn]; 27 int ans; 28 29 void dfs1(int now){ 30 for(int i=0;i<ve[now].size();i++){ 31 parent[ve[now][i]][0]=now; 32 val[ve[now][i]][0]=a[now]; 33 dfs1(ve[now][i]); 34 } 35 } 36 37 ll dfs2(int now){ 38 ll sum=0; 39 for(int i=0;i<ve[now].size();i++){ 40 sum=max(sum,dfs2(ve[now][i])); 41 } 42 if(sum==0){ 43 ans++; 44 return len[now]-1; 45 } 46 return sum-1; 47 } 48 49 int main(){ 50 cin>>n>>L>>S; 51 for(int i=1;i<=n;i++){ 52 cin>>a[i]; 53 if(a[i]>S){ 54 cout<<-1<<endl; 55 return 0; 56 } 57 } 58 int x; 59 for(int i=2;i<=n;i++){ 60 cin>>x; 61 ve[x].pb(i); 62 } 63 dfs1(1); 64 for(int i=1;i<=18;i++){ 65 for(int j=1;j<=n;j++){ 66 parent[j][i]=parent[parent[j][i-1]][i-1]; 67 val[j][i]=val[parent[j][i-1]][i-1]+val[j][i-1]; 68 } 69 } 70 for(int i=1;i<=n;i++){ 71 ll sum=a[i],now=i; 72 len[i]=1; 73 for(int j=18;j>=0&&sum<=S&&len[i]<=L;j--){ 74 if(parent[now][j]&&sum+val[now][j]<=S&&len[i]+(1<<j)<=L){ 75 sum+=val[now][j]; 76 now=parent[now][j]; 77 len[i]+=(1<<j); 78 } 79 } 80 } 81 dfs2(1); 82 cout<<ans<<endl; 83 }