D. Multiple Testcases（贪心、dp Educational Codeforces Round 86 (Rated for Div. 2)）

D. Multiple Testcases

CF 1342D I'm so
angry, I'm just a ceil, my god, wsm is always like this, I vomit, I'm so sad

Meaning:
Given $n,k$ , given$n$ arrays$m_i$, Request to group them.
Given $k$ size array$c_i$, Requires that each group is greater than or equal to iiThe number of digits of$i$ is less than or equal to$c_i$.
Find the minimum number of groups anss, and output the numbers allocated in each group.

Idea:
I think about it at the beginning, there is only one differenceceil
First find the minimum grouping number anss: for $m$ array sorting,$d{p}_i$Record greater than or equal to iiThe number of$i$ , the current anss is$ceil(d{p}_i/c_i)$ , Take the maximum value of anss

int n;
vector<LL> ans[maxn];
LL m[maxn],c[maxn],dp[maxn],mp[maxn],anss;
int main(){
    
    
    LL n=lrd(),k=lrd();
    anss=0;
    for(int i=1;i<=n;i++){
    
    m[i]=lrd();mp[m[i]]++;}
    sort(m+1,m+1+n);
    for(int i=k;i>=1;i--)dp[i]=dp[i+1]+mp[i];
    for(int i=1;i<=k;i++){
    
    c[i]=ird();anss=max(anss,(dp[i]+c[i]-1)/c[i]);}
    cout<<anss<<endl;
    int nw=0;
    for(int i=n;i>=1;i--){
    
    
        ans[nw].push_back(m[i]);
        nw++;
        nw%=anss;
    }
    for(int i=0;i<anss;i++){
    
    
        cout<<ans[i].size()<<" ";
        for(int j=0;j<ans[i].size();j++)
            cout<<ans[i][j]<<" ";
        cout<<endl;
    }
}