In order to solve this problem, we must make a series of observations. The first is that it doesn’t matter what order the rhyme classes are given - what matters is the frequencies of each rhyme class. Once we compute the frequency of each rhyme class $f_i$ and the number of ways to end a line with a rhyme $w_k,$ the number of poems that bessie can write for that rhyme class is

For instance, there are $8$ ways to end in rhyme $1$ and $4$ ways to end in rhyme $2$. There are $2$ of rhyme class $A$ and $1$ of rhyme class $B$. Thus, the answer is

Now, we need to find the number of ways to form a line that ends in each rhyme. To do this, we use dynamic programming. Let $dp[i]$ be the number of ways to form a line of length $i.$ The transition between states is, for the number of syllables in every word $w_s,$ is:

(Don’t forget to modulo 1,000,000,007 at every step!)

Whenever $w_s + i == K,$ we increment $r[w_r],$ where $w_r$ is the rhyme of that word and $r$ is an array storing the number of ways to form a line that ends in each rhyme.

Finally, to implement the math formula that we discussed in the first paragraph, we need to use an efficient exponentiation algorithm, which can compute $E^{P} \mod 1000000007$ in $O(\log{P}).$ Using std::pow() or Math.pow() will not work because of the possibility of large numbers.

The efficiency of the dynamic programming is $O(NK)$ and the exponentiation is $O(NM\log{M} )$ for the worst case, which fits inside the time limit.

#include <bits/stdc++.h>

#define MAXN 5005
#define MAXM 100005
#define MAXS 5005


using namespace std;

long long MOD = 1000000007;

//fast exponentiation
long long exp(int base, int power){
   if(power == 0) return 1;
   if(power == 1) return (base + MOD) % MOD;
   long long ans = exp(base,power/2);
   ans = (ans * ans + MOD) % MOD;
   if(power%2 == 1) ans = (ans*base + MOD) % MOD;
   return (ans + MOD) % MOD;
}

int n,m,s;
long long dp[MAXS];                       
//dp[x] = the number of ways to make a line with x syllables.
long long r[MAXN];                        
//r[x] = the number of ways to form a full line that ends with rhyme scheme x

unordered_map<char,int> umap;

int main(){
   ios::sync_with_stdio(false);
   cin.tie(0);
   
   ifstream fin ("poetry.in");
   ofstream fout ("poetry.out");
   
   int n,m,s;
   fin >> n >> m >> s;
   
   pair<int,int> words[n];          //first is syllables, second is rhyme
   
   for(int k = 0; k < n; k++){
      int a,b;
      fin >> a >> b;
      words[k] = make_pair(a,b);
   }
   
   //Calculate frequencies of every rhyme (Order of rhymes doesn't matter)
   for(int k = 0; k < m; k++){
      char c;
      fin >> c;
      if(umap.find(c) == umap.end()){
         umap[c] = 1;
      } else{
         umap[c]++;
      }
   }
   
   
   dp[0] = 1;
   
   for(int k = 0; k <= s; k++){
      
      for(int j = 0; j < n; j++){
         if(words[j].first + k > s) continue;
         if(words[j].first + k == s){
            r[words[j].second] = (r[words[j].second] + dp[k] + MOD) % MOD;                      //if you are at the end of the line, update r
         } else {
            dp[words[j].first + k] = (dp[words[j].first + k] + dp[k] + MOD) % MOD;              //knapsack dp
         }
      }
   }
   
   long long answer = 1;
   for(auto a : umap){
      //use counting/probability to calculate the answer. 
      //For every grouping of a rhyme, multiply the answer by 
      //r[1]^freq, r[2]^freq, etc.
      //For instance, the answer for the sample case is 
      //(8^2 + 4^2) * (8^1 + 4^1). r[1] = 8 and r[2] = 4, and the are 2 As and 1 B.
      
   
      int freq = a.second;
      long long sum = 0;
      for(int k = 0; k <= n; k++){
         if(r[k] == 0) continue;
         sum = (sum + exp(r[k],freq) + MOD) % MOD;
      }
      answer = (answer * sum + MOD) % MOD;
   }
   
   cout << answer;
   fout << answer;
         
   return 0;
}

Here is the same solution but in Java:

import java.io.*;
import java.util.*;

class poetry{
   
   public static long MOD = 1000000007L;
   
   public static void main(String[] args) throws IOException{
      BufferedReader f = new BufferedReader(new FileReader("poetry.in"));
      PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("poetry.out")));
      
      StringTokenizer st = new StringTokenizer(f.readLine());
      
      int n = Integer.parseInt(st.nextToken());
      int m = Integer.parseInt(st.nextToken());
      int s = Integer.parseInt(st.nextToken());
      
      Word[] words = new Word[n];
      
      for(int k = 0; k < n; k++){
         
         st = new StringTokenizer(f.readLine());
      
         int sy = Integer.parseInt(st.nextToken());
         int rh = Integer.parseInt(st.nextToken());
      
         words[k] = new Word(sy,rh);
      }
      
      
      //Calculate frequencies of every rhyme (Order of rhymes doesn't matter)
      HashMap<Character,Integer> hmap = new HashMap<Character,Integer>();
      
      
      for(int k = 0; k < m; k++){
         char c = f.readLine().charAt(0);
         if(hmap.containsKey(c)){
            hmap.put(c,hmap.get(c)+1);
         } else {
            hmap.put(c,1);
         }
      }
      
      //dp[x] = the number of ways to make a line with x syllables.
      long[] dp = new long[s+1];
      dp[0] = 1L;
      
      //r[x] = the number of ways to form a full line that ends with rhyme scheme x
      long[] r = new long[n+1];
      
      for(int k = 0; k <= s; k++){
         
         for(int j = 0; j < n; j++){
            if(words[j].s + k > s) continue;
            if(words[j].s + k == s){
               r[words[j].r] = (r[words[j].r] + dp[k] + MOD) % MOD;                 //if you are at the end of the line, update r
            }
            dp[words[j].s + k] = (dp[words[j].s + k] + dp[k] + MOD) % MOD;          //knapsack dp
         }
      }      
      
      
      long answer = 1L;
      for(char c : hmap.keySet()){
         //use counting/probability to calculate the answer. For every grouping of a rhyme, multiple the answer by r[1]^freq, r[2]^freq, etc.
         //For instance, the answer for the sample case is (8^2 + 4^2) * (8^1 + 4^1). r[1] = 8 and r[2] = 4, and the are 2 As and 1 B.
      
         int freq = hmap.get(c);
         long sum = 0L;
         for(int k = 0; k < r.length; k++){
            if(r[k] == 0) continue;
            sum = (sum + exp(r[k],freq) + MOD) % MOD;
         }
      
         answer = (answer * sum + MOD) % MOD;
      }
      
      System.out.println(answer);
      out.println(answer);
      
         
      
      
      
        
      out.close();
   }
   
   
   //fast exponentiation
   public static long exp(long base, int power){
      if(power == 0) return 1;
      if(power == 1) return (base + MOD) % MOD;
      long ans = exp(base,power/2);
      ans = (ans*ans + MOD) % MOD;
      if(power%2 == 1) ans = (ans*base + MOD) % MOD;
      return (ans + MOD) % MOD;
   }
   
   public static class Word{
      int s;                     //syllables
      int r;                     //rhyme
      public Word(int a, int b){
         s = a;
         r = b;
      }
   }              
}