The Preliminary Contest for ICPC Asia Shanghai 2019 F. Rhyme scheme(dp)

 

 

 

 Meaning of the questions: to give you a n and k want you to find a sequence of length n k dictionary definition of a small string is legal: the i range of characters can only be before i-1 character maximum value +1

Ideas: we dp [n] [i] [j] represents the length of the bit i n j maximum value of the number of sequence and then we can find a direct imitation of the k-k-found big as string

#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const double eps = 1e-6;
const int N = 1e6+7;
typedef long long ll;
typedef __int128 bll;
const ll mod = 998244353;
using namespace std;
inline __int128 read() {
     __int128 x=0,f=1;
     char ch=getchar();
     while(ch<'0'||ch>'9') {
         if(ch=='-')
             f=-1;
         ch=getchar();
     }
     while(ch>='0'&&ch<='9') {
         x=x*10+ch-'0';
         ch=getchar();
     }
     return x*f;
}
inline void print(__int128 x)
{    
   if(x<0){putchar('-');x=-x;}
   if(x>9) print(x/10);
   putchar(x%10+'0');
}
bll dp[30][30][30];
char v[30];
bll dfs(int len,int now,int mx){
    if(now==len){
        dp[len][now][mx]=1;
        return dp[len][now][mx];
    }
    if(dp[len][now][mx]!=-1) return dp[len][now][mx];
    bll ans=0;
    for(int i=1;i<=min(mx+1,26);i++){
        if(i<=mx){
            ans+=dfs(len,now+1,mx);
        }else{
            ans+=dfs(len,now+1,i);
        }
    }
    dp[len][now][mx]=ans;
    return ans;
}
int main(){
//    ios::sync_with_stdio(false);
//    cin.tie(0); cout.tie(0);
    int t; scanf("%d",&t);
    for(int i=0;i<30;i++)
        for(int j=0;j<30;j++)
            for(int k=0;k<30;k++)
                dp[i][j][k]=-1;
    for(int i=1;i<=26;i++)
        dfs(i,1,1);
    int w=0;
//    print(dp[3][1][1]);
    while(t--){
        int n; scanf("%d",&n);
        bll k; k=read();
        int mx=1;
        printf("Case #%d: ",++w);
        for(int i=1;i<=n;i++){
            v[i]='A'; 
            for(int j=1;j<=mx+1;j++){
                //cout<<k<<" "<<dp[n][i][j]<<endl;
                int p=max(mx,j);
                if(dp[n][i][p]>=k){
                //    mx=max(mx,j);
                    v[i]=char(j+'A'-1);
                    //putchar('A' + j - 1);
                    mx=max(mx,p);
                //    cout<<j<<endl;
                    break;
                }else{
                    k-=dp[n][i][p];
                }
            }
        }
        for(int i=1;i<=n;i++)
            printf("%c",v[i]);
        puts("");
    }
    return 0;
}

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