Codeforces Round #518 (Div. 2) [Thanks, Mail.Ru!]

Codeforces Round #518 (Div. 2) [Thanks, Mail.Ru!]

https://codeforces.com/contest/1068

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 300005
 #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 pair<int,int> pii;
 typedef pair<long long,int>pli;
 typedef pair<int,char> pic;
 typedef pair<pair<int,string>,pii> ppp;
 typedef unsigned long long ull;
 const long long mod=1e9+;
 /*#ifndef ONLINE_JUDGE
         freopen("1.txt","r",stdin);
 #endif */

 int main(){
     #ifndef ONLINE_JUDGE
      //   freopen("1.txt","r",stdin);
     #endif
     std::ios::sync_with_stdio(false);
     ll n,m,k,l;
     cin>>n>>m>>k>>l;
     if(n<m||n-k<l){
            cout<<-<<endl;
         return ;
     }
     ll res=(k+l)/m;
     if(res*m<k+l) ++res;
     if(res*m<=n) cout<<res<<endl;
     else cout<<-<<endl;
 }

B

数论

因为lcm(a,b)/a==b/gcd(a,b),又因为b是确定的，所以求的是gcd(a,b)的个数，也就是求b的因子数

 #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 300005
 #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 pair<int,int> pii;
 typedef pair<long long,int>pli;
 typedef pair<int,char> pic;
 typedef pair<pair<int,string>,pii> ppp;
 typedef unsigned long long ull;
 const long long mod=1e9+;
 /*#ifndef ONLINE_JUDGE
         freopen("1.txt","r",stdin);
 #endif */

 int main(){
     #ifndef ONLINE_JUDGE
      //   freopen("1.txt","r",stdin);
     #endif
     std::ios::sync_with_stdio(false);
     ll n;
     cin>>n;
     int sq=sqrt(n);
     ll ans=;
     for(int i=;i<=sq;i++){
         ll co=;
         while(n%i==){
             n/=i;
             co++;
         }
         ans*=co;
     }
     if(n>) ans*=;
     cout<<ans<<endl;
 }

C

找规律

 #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 300005
 #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 pair<int,int> pii;
 typedef pair<long long,int>pli;
 typedef pair<int,char> pic;
 typedef pair<pair<int,string>,pii> ppp;
 typedef unsigned long long ull;
 const long long mod=1e9+;
 /*#ifndef ONLINE_JUDGE
         freopen("1.txt","r",stdin);
 #endif */

 vector<int>ve[];
 int n,m;

 int main(){
     #ifndef ONLINE_JUDGE
      //   freopen("1.txt","r",stdin);
     #endif
     std::ios::sync_with_stdio(false);
     cin>>n>>m;
     for(int i=;i<=n;i++){
         ve[i].pb(i);
     }
     int x,y;
     int co=n+;
     for(int i=;i<m;i++){
         cin>>x>>y;
         ve[x].pb(co);
         ve[y].pb(co);
         co++;
     }
     for(int i=;i<=n;i++){
         cout<<ve[i].size()<<endl;
         for(int j=;j<ve[i].size();j++){
             cout<<i<<" "<<ve[i][j]<<endl;
         }
     }
 }

D

DP

参考博客：https://blog.csdn.net/white_156/article/details/83421537

 #include<iostream>
 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 100005
 #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 pair<int,int> pii;
 typedef pair<long long,int>pli;
 typedef pair<int,char> pic;
 typedef pair<pair<int,string>,pii> ppp;
 typedef unsigned long long ull;
 const long long mod=;
 /*#ifndef ONLINE_JUDGE
         freopen("1.txt","r",stdin);
 #endif */

 int a[];
 long long dp[][][];

 int main(){
     #ifndef ONLINE_JUDGE
      //   freopen("1.txt","r",stdin);
     #endif
     std::ios::sync_with_stdio(false);
     int n;
     scanf("%d",&n);
     for(int i=;i<=n;i++) cin>>a[i];
     if(a[]!=-){
         dp[][a[]][]=;
     }
     else{
         for(int i=;i<=;i++){
             dp[][i][]=;
             dp[][i][]=;
         }
     }
     for(int i=;i<=n;i++){
         if(a[i]==-){
             dp[i][][]=;
             for(int j=;j<=;j++)
                 dp[i][j][]=(dp[i][j][]+dp[i][j-][]+dp[i-][j-][]+dp[i-][j-][])%mod;
             dp[i][][]=;
             for(int j=;j>=;j--){
                 dp[i][j][]=(dp[i][j][]+dp[i][j+][]+dp[i-][j+][])%mod;
             }
             for(int j=;j<=;j++)
                 dp[i][j][]=(dp[i][j][]+dp[i-][j][]+dp[i-][j][])%mod;
         }
         else{
             int num=a[i];
             for(int j=;j<num;j++)
                 dp[i][num][]=(dp[i][num][]+dp[i-][j][]+dp[i-][j][])%mod;
             for(int j=num+;j<=;j++)
                 dp[i][num][]=(dp[i][num][]+dp[i-][j][])%mod;
             dp[i][num][]=(dp[i][num][]+dp[i-][num][]+dp[i-][num][])%mod;
         }
     }
     long long ans=;
     for(int i=;i<=;i++)
         ans=(ans+dp[n][i][])%mod;
     cout<<ans<<endl;

 }

E

模拟

 #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 100005
 #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 pair<int,int> pii;
 typedef pair<long long,int>pli;
 typedef pair<int,char> pic;
 typedef pair<pair<int,string>,pii> ppp;
 typedef unsigned long long ull;
 const long long mod=;
 /*#ifndef ONLINE_JUDGE
         freopen("1.txt","r",stdin);
 #endif */

 set<int>se[maxn],v,t;
 set<int>::iterator it;
 int cnt[maxn];

 int main(){
     #ifndef ONLINE_JUDGE
      //   freopen("1.txt","r",stdin);
     #endif
     std::ios::sync_with_stdio(false);
     int n,k;
     cin>>n>>k;
     for(int i=;i<n;i++)
     {
         int u,v;
         cin>>u>>v;
         se[u].insert(v);
         se[v].insert(u);
     }
     for(int i=;i<=n;i++)
         if(se[i].size()==)
             v.insert(i);
     while(n>)
     {
         t.clear();
         for(it=v.begin();it!=v.end();it++)
         {
             int x=*se[*it].begin();
             cnt[x]++;
             t.insert(x);
             se[x].erase(*it);
             n--;
         }
         for(it=t.begin();it!=t.end();it++)
         {
             if(cnt[*it]<)
             {
                 printf("No\n");
                 return ;
             }
             cnt[*it]=;
         }
         swap(v,t);
         k--;
     }
     if(k==)printf("Yes\n");
     else printf("No\n");

 }