kickstart 2020 Round A

Well boring (((
recommended kickstart renamed div4, of course, I just suggest
for the first time to play, he would not resort to the
A:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef double db;
const int N = 3e5+5;
int t,n,a[100005],b,cas;
int main(){
    ios::sync_with_stdio(false);
    cin>>t;
    while (t--){
        cas++;
        cin>>n>>b;
        for(int i=1;i<=n;i++)cin>>a[i];
        sort(a+1,a+1+n);
        int now=0;
        while (now<n&&b>=a[now+1]){
            b-=a[now+1];
            now++;
        }
        cout<<"Case #"<<cas<<": ";
        cout<<now<<endl;
    }
}

B:
dp [i] [j] represents the j-th selected answers before i pile items up to the next election to enumerate the number of pile

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef double db;
const int N = 3e5+5;
int t,n,k,p;
int v[55][55];
int dp[55][1555],f[55][55];
int main(){
    ios::sync_with_stdio(false);
    cin>>t;
    int cas=0;
    while (t--){
        cas++;
        memset(dp,0, sizeof(dp));
        cin>>n>>k>>p;
        for(int i=1;i<=n;i++)for(int j=1;j<=k;j++)cin>>v[i][j],f[i][j]=v[i][j]+f[i][j-1];
        for(int i=1;i<=n;i++){
            for(int j=p;j>=0;j--){
                for(int s=0;s<=j&&s<=k;s++){
                    dp[i][j]=max(dp[i][j],dp[i-1][j-s]+f[i][s]);
                }
            }
        }
        cout<<"Case #"<<cas<<": ";
        cout<<dp[n][p]<<endl;
    }
}

C:
half

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef double db;
const int N = 1e5+5;
int t,n,k,a[N];
bool can(int x){
    int ned = 0;
    for(int i=2;i<=n;i++){
        int dis = a[i]-a[i-1];
        int c = dis/x;
        if(dis%x==0)c--;
        ned+=c;
    }
    return ned<=k;
}
int main(){
    ios::sync_with_stdio(false);
    cin>>t;
    int cas = 0;
    while (t--){
        cas++;
        cin>>n>>k;
        for(int i=1;i<=n;i++)cin>>a[i];
        int l=1,r=1e9;
        while (l<=r){
            int mid = l+r>>1;
            if(can(mid)){
                r=mid-1;
            }else{
                l=mid+1;
            }
        }
        cout<<"Case #"<<cas<<": ";
        cout<<l<<endl;
    }
}

D:
I think for a few minutes, the first thought may be greedy if selected, is always looking for the longest common prefix string a bunch of out
it is easy to find this greed is right
then that sucker trie tree, the time complexity is also very good seeking
the way this evaluation Kyi really powerful, I memset actually all right. . .

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5+5;
int t,n,k;
string s[N];
int tree[5000000][26],pre[5000000],tot;
int sum[5000000];
int cas = 0;
void insert(string str){
    int root = 0;
    for(int i=0;i<str.length();i++){
        int id = str[i]-'A';
        if(!tree[root][id])
            tree[root][id]=++tot;
        sum[tree[root][id]]++;
        pre[tree[root][id]]=root;
        root = tree[root][id];
    }
}
void slove(){
    cas++;
    int ans = 0;
    for(int i=tot;i>=1;i--){
        while (sum[i]>=k){
            int dep=0;
            int now = i;
            while (now!=0){
                sum[now]-=k;
                dep++;
                now=pre[now];
            }
            ans+=dep;
        }
    }
    cout<<"Case #"<<cas<<": ";
    cout<<ans<<endl;
}
int main(){
    ios::sync_with_stdio(false);
    cin>>t;
    while (t--){
        cin>>n>>k;// n/k
        for(int i=1;i<=n;i++)cin>>s[i],insert(s[i]);
        slove();
        memset(pre,0, sizeof(pre));
        memset(sum,0, sizeof(sum));
//        for(int i=0;i<=tot;i++)pre[tot]=sum[tot]=0;
        tot=0;
        memset(tree,0, sizeof(tree));
    }
}