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


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

    https://codeforces.com/contest/1068

    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 300005
     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 /*#ifndef ONLINE_JUDGE
    20         freopen("1.txt","r",stdin);
    21 #endif */
    22 
    23 
    24 
    25 
    26 int main(){
    27     #ifndef ONLINE_JUDGE
    28      //   freopen("1.txt","r",stdin);
    29     #endif
    30     std::ios::sync_with_stdio(false);
    31     ll n,m,k,l;
    32     cin>>n>>m>>k>>l;
    33     if(n<m||n-k<l){
    34            cout<<-1<<endl;
    35         return 0;
    36     }
    37     ll res=(k+l)/m; 
    38     if(res*m<k+l) ++res; 
    39     if(res*m<=n) cout<<res<<endl;
    40     else cout<<-1<<endl;
    41 }
    View Code

    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

     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 300005
     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 /*#ifndef ONLINE_JUDGE
    20         freopen("1.txt","r",stdin);
    21 #endif */
    22 
    23 
    24 
    25 
    26 int main(){
    27     #ifndef ONLINE_JUDGE
    28      //   freopen("1.txt","r",stdin);
    29     #endif
    30     std::ios::sync_with_stdio(false);
    31     ll n;
    32     cin>>n;
    33     int sq=sqrt(n);
    34     ll ans=1;
    35     for(int i=2;i<=sq;i++){
    36         ll co=1;
    37         while(n%i==0){
    38             n/=i;
    39             co++;
    40         }
    41         ans*=co;
    42     }
    43     if(n>1) ans*=2;
    44     cout<<ans<<endl;
    45 }
    View Code

    C

    找规律

     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 300005
     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 /*#ifndef ONLINE_JUDGE
    20         freopen("1.txt","r",stdin);
    21 #endif */
    22 
    23 
    24 vector<int>ve[105];
    25 int n,m;
    26 
    27 int main(){
    28     #ifndef ONLINE_JUDGE
    29      //   freopen("1.txt","r",stdin);
    30     #endif
    31     std::ios::sync_with_stdio(false);
    32     cin>>n>>m;
    33     for(int i=1;i<=n;i++){
    34         ve[i].pb(i);
    35     }
    36     int x,y;
    37     int co=n+1;
    38     for(int i=0;i<m;i++){
    39         cin>>x>>y;
    40         ve[x].pb(co);
    41         ve[y].pb(co);
    42         co++;
    43     }
    44     for(int i=1;i<=n;i++){
    45         cout<<ve[i].size()<<endl;
    46         for(int j=0;j<ve[i].size();j++){
    47             cout<<i<<" "<<ve[i][j]<<endl;
    48         }
    49     }
    50 }
    View Code

    D

    DP

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

     1 #include<iostream>
     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<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=998244353;
    19 /*#ifndef ONLINE_JUDGE
    20         freopen("1.txt","r",stdin);
    21 #endif */
    22 
    23 int a[100005];
    24 long long dp[100005][205][2];
    25 
    26 
    27 int main(){
    28     #ifndef ONLINE_JUDGE
    29      //   freopen("1.txt","r",stdin);
    30     #endif
    31     std::ios::sync_with_stdio(false);
    32     int n;
    33     scanf("%d",&n);
    34     for(int i=1;i<=n;i++) cin>>a[i];
    35     if(a[1]!=-1){
    36         dp[1][a[1]][0]=1;
    37     }
    38     else{
    39         for(int i=1;i<=200;i++){
    40             dp[1][i][0]=1;
    41             dp[1][i][1]=0;
    42         }
    43     }
    44     for(int i=2;i<=n;i++){
    45         if(a[i]==-1){
    46             dp[i][1][0]=0;
    47             for(int j=2;j<=200;j++)
    48                 dp[i][j][0]=(dp[i][j][0]+dp[i][j-1][0]+dp[i-1][j-1][0]+dp[i-1][j-1][1])%mod;
    49             dp[i][200][1]=0;
    50             for(int j=199;j>=1;j--){
    51                 dp[i][j][1]=(dp[i][j][1]+dp[i][j+1][1]+dp[i-1][j+1][1])%mod;
    52             }
    53             for(int j=1;j<=200;j++)
    54                 dp[i][j][1]=(dp[i][j][1]+dp[i-1][j][1]+dp[i-1][j][0])%mod;
    55         }
    56         else{
    57             int num=a[i];
    58             for(int j=1;j<num;j++)
    59                 dp[i][num][0]=(dp[i][num][0]+dp[i-1][j][0]+dp[i-1][j][1])%mod;
    60             for(int j=num+1;j<=200;j++)
    61                 dp[i][num][1]=(dp[i][num][1]+dp[i-1][j][1])%mod;
    62             dp[i][num][1]=(dp[i][num][1]+dp[i-1][num][0]+dp[i-1][num][1])%mod;
    63         }
    64     }
    65     long long ans=0;
    66     for(int i=1;i<=200;i++)
    67         ans=(ans+dp[n][i][1])%mod;
    68     cout<<ans<<endl;
    69 
    70     
    71     
    72 }
    View Code

    E

    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 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<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=998244353;
    19 /*#ifndef ONLINE_JUDGE
    20         freopen("1.txt","r",stdin);
    21 #endif */
    22 
    23 set<int>se[maxn],v,t;
    24 set<int>::iterator it;
    25 int cnt[maxn];
    26 
    27 
    28 int main(){
    29     #ifndef ONLINE_JUDGE
    30      //   freopen("1.txt","r",stdin);
    31     #endif
    32     std::ios::sync_with_stdio(false);
    33     int n,k;
    34     cin>>n>>k;
    35     for(int i=1;i<n;i++)
    36     {
    37         int u,v;
    38         cin>>u>>v;
    39         se[u].insert(v);
    40         se[v].insert(u);
    41     }
    42     for(int i=1;i<=n;i++)
    43         if(se[i].size()==1)
    44             v.insert(i);
    45     while(n>1)
    46     {
    47         t.clear();
    48         for(it=v.begin();it!=v.end();it++)
    49         {
    50             int x=*se[*it].begin();
    51             cnt[x]++;
    52             t.insert(x);
    53             se[x].erase(*it);
    54             n--;
    55         }
    56         for(it=t.begin();it!=t.end();it++)
    57         {
    58             if(cnt[*it]<3)
    59             {
    60                 printf("No
    ");
    61                 return 0;
    62             }
    63             cnt[*it]=0;
    64         }
    65         swap(v,t);
    66         k--;
    67     }
    68     if(k==0)printf("Yes
    ");
    69     else printf("No
    ");
    70 
    71     
    72     
    73 }
    View Code
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  • 原文地址:https://www.cnblogs.com/Fighting-sh/p/10550568.html
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