最多 K 个连续元音的大小为 N 的不同单词的数量

给定两个整数N和K ，任务是找到由长度为N的小写字母组成的不同字符串的数量，这些字母最多可以由K个连续元音组成。由于答案可能太大，打印answer%1000000007 。

Input: N = 1, K = 0
Output: 21
Explanation: All the 21 consonants are there which has 0 contiguous vowels and are of length 1.

Input: N = 1, K = 1
Output: 26

方法：解决这个问题的思路是基于动态规划。请按照以下步骤解决问题：

• 让DP [i] [j]是的办法，使长度i的不同的字符串，其中字符串的最后Ĵ字符是元音数量。
• 所以动态规划的状态是：
  • 如果j = 0 ，则dp[i][j] = (dp[i-1][0] + dp[i-1][1] +……+ dp[i-1][K])*21 (由整数变量sum 表示) 因为最后添加的字符应该是辅音，而不管它在先前状态中的值如何，只有j的值将变为0 。
  • 如果i则dp[i][j] = 0 。因为不可能创建一个包含j个元音并且长度小于j的字符串。
  • 如果我== j时，DP [i] [j] = 5 ^I，因为在字符串中的字符数等于元音的数量，因此所有字符应为元音。
  • 如果j 那么DP [i] [j] = DP [I-1] [j-1] * 5，因为长度i与去年Ĵ字符的元音的字符串可以由仅当最后一个字符是元音和长度为i-1 的字符串将最后j – 1 个字符作为元音。
• 打印dp[n][0] + dp[n][1] + …… + dp[n][K] 之和作为答案。

下面是上述方法的实现

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Power function to calculate
// long powers with mod
long long int power(long long int x,
                    long long int y,
                    long long int p)
{
    long long int res = 1ll;
 
    x = x % p;
 
    if (x == 0)
        return 0;
 
    while (y > 0) {
 
        if (y & 1)
            res = (res * x) % p;
        y = y >> 1;
        x = (x * x) % p;
    }
    return res;
}
 
// Function for finding number of ways to
// create string with length N and atmost
// K contiguous vowels
int kvowelwords(int N, int K)
{
 
    long long int i, j;
    long long int MOD = 1000000007;
 
    // Array dp to store number of ways
    long long int dp[N + 1][K + 1] = { 0 };
 
    long long int sum = 1;
    for (i = 1; i <= N; i++) {
 
        // dp[i][0] = (dp[i-1][0]+dp[i-1][1]..dp[i-1][k])*21
        dp[i][0] = sum * 21;
        dp[i][0] %= MOD;
 
        // Now setting sum to be dp[i][0]
        sum = dp[i][0];
 
        for (j = 1; j <= K; j++) {
            // If j>i, no ways are possible to create
            // a string with length i and vowel j
            if (j > i)
                dp[i][j] = 0;
            else if (j == i) {
                // If j = i all the character should
                // be vowel
                dp[i][j] = power(5ll, i, MOD);
            }
            else {
                // dp[i][j] relation with dp[i-1][j-1]
                dp[i][j] = dp[i - 1][j - 1] * 5;
            }
 
            dp[i][j] %= MOD;
 
            // Adding dp[i][j] in the sum
            sum += dp[i][j];
            sum %= MOD;
        }
    }
 
    return sum;
}
// Driver Program
int main()
{
    // Input
    int N = 3;
    int K = 3;
 
    // Function Call
    cout << kvowelwords(N, K) << endl;
    return 0;
}

Java

// Java program for the above approach
class GFG{
 
// Power function to calculate
// long powers with mod
static int power(int x, int y, int p)
{
    int res = 1;
    x = x % p;
  
    if (x == 0)
        return 0;
  
    while (y > 0)
    {
        if ((y & 1) != 0)
            res = (res * x) % p;
             
        y = y >> 1;
        x = (x * x) % p;
    }
    return res;
}
  
// Function for finding number of ways to
// create string with length N and atmost
// K contiguous vowels
static int kvowelwords(int N, int K)
{
    int i, j;
    int MOD = 1000000007;
  
    // Array dp to store number of ways
    int[][] dp = new int[N + 1][K + 1] ;
  
    int sum = 1;
    for(i = 1; i <= N; i++)
    {
         
        // dp[i][0] = (dp[i-1][0]+dp[i-1][1]..dp[i-1][k])*21
        dp[i][0] = sum * 21;
        dp[i][0] %= MOD;
  
        // Now setting sum to be dp[i][0]
        sum = dp[i][0];
  
        for(j = 1; j <= K; j++)
        {
             
            // If j>i, no ways are possible to create
            // a string with length i and vowel j
            if (j > i)
                dp[i][j] = 0;
                 
            else if (j == i)
            {
                 
                // If j = i all the character should
                // be vowel
                dp[i][j] = power(5, i, MOD);
            }
            else
            {
                 
                // dp[i][j] relation with dp[i-1][j-1]
                dp[i][j] = dp[i - 1][j - 1] * 5;
            }
  
            dp[i][j] %= MOD;
  
            // Adding dp[i][j] in the sum
            sum += dp[i][j];
            sum %= MOD;
        }
    }
    return sum;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Input
    int N = 3;
    int K = 3;
  
    // Function Call
    System.out.println( kvowelwords(N, K));
}
}
 
// This code is contributed by target_2

Python3

# Python3 program for the above approach
 
# Power function to calculate
# long powers with mod
def power(x, y, p):
     
    res = 1
    x = x % p
 
    if (x == 0):
        return 0
 
    while (y > 0):
        if (y & 1):
            res = (res * x) % p
             
        y = y >> 1
        x = (x * x) % p
         
    return res
 
# Function for finding number of ways to
# create string with length N and atmost
# K contiguous vowels
def kvowelwords(N, K):
 
    i, j = 0, 0
    MOD = 1000000007
 
    # Array dp to store number of ways
    dp = [[0 for i in range(K + 1)]
             for i in range(N + 1)]
 
    sum = 1
    for i in range(1, N + 1):
         
        #dp[i][0] = (dp[i-1][0]+dp[i-1][1]..dp[i-1][k])*21
        dp[i][0] = sum * 21
        dp[i][0] %= MOD
 
        # Now setting sum to be dp[i][0]
        sum = dp[i][0]
 
        for j in range(1, K + 1):
             
            # If j>i, no ways are possible to create
            # a string with length i and vowel j
            if (j > i):
                dp[i][j] = 0
            elif (j == i):
                 
                # If j = i all the character should
                # be vowel
                dp[i][j] = power(5, i, MOD)
            else:
                 
                # dp[i][j] relation with dp[i-1][j-1]
                dp[i][j] = dp[i - 1][j - 1] * 5
 
            dp[i][j] %= MOD
 
            # Adding dp[i][j] in the sum
            sum += dp[i][j]
            sum %= MOD
 
    return sum
     
# Driver Code
if __name__ == '__main__':
     
    # Input
    N = 3
    K = 3
 
    # Function Call
    print (kvowelwords(N, K))
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Power function to calculate
// long powers with mod
static int power(int x, int y, int p)
{
    int res = 1;
    x = x % p;
  
    if (x == 0)
        return 0;
  
    while (y > 0)
    {
        if ((y & 1) != 0)
            res = (res * x) % p;
             
        y = y >> 1;
        x = (x * x) % p;
    }
    return res;
}
  
// Function for finding number of ways to
// create string with length N and atmost
// K contiguous vowels
static int kvowelwords(int N, int K)
{
    int i, j;
    int MOD = 1000000007;
  
    // Array dp to store number of ways
    int[,] dp = new int[N + 1, K + 1];
  
    int sum = 1;
    for(i = 1; i <= N; i++)
    {
         
        // dp[i][0] = (dp[i-1, 0]+dp[i-1, 1]..dp[i-1][k])*21
        dp[i, 0] = sum * 21;
        dp[i, 0] %= MOD;
  
        // Now setting sum to be dp[i][0]
        sum = dp[i, 0];
  
        for(j = 1; j <= K; j++)
        {
             
            // If j>i, no ways are possible to create
            // a string with length i and vowel j
            if (j > i)
                dp[i, j] = 0;
                 
            else if (j == i)
            {
                 
                // If j = i all the character should
                // be vowel
                dp[i, j] = power(5, i, MOD);
            }
            else
            {
                 
                // dp[i][j] relation with dp[i-1][j-1]
                dp[i, j] = dp[i - 1, j - 1] * 5;
            }
  
            dp[i, j] %= MOD;
  
            // Adding dp[i][j] in the sum
            sum += dp[i, j];
            sum %= MOD;
        }
    }
    return sum;
}
 
// Driver Code
public static void Main()
{
     
    // Input
    int N = 3;
    int K = 3;
  
    // Function Call
    Console.Write(kvowelwords(N, K));
}
}
 
// This code is contributed by code_hunt

Javascript

 
// JavaScript code for above approach
 
// Power function to calculate
// long powers with mod
function power(x, y, p)
{
    let res = 1;
    x = x % p;
  
    if (x == 0)
        return 0;
  
    while (y > 0)
    {
        if ((y & 1) != 0)
            res = (res * x) % p;
             
        y = y >> 1;
        x = (x * x) % p;
    }
    return res;
}
  
// Function for finding number of ways to
// create string with length N and atmost
// K contiguous vowels
function kvowelwords(N, K)
{
    let i, j;
    let MOD = 1000000007;
  
    // Array dp to store number of ways
    let dp = new Array(N + 1)
    // Loop to create 2D array using 1D array
    for (i = 0; i < dp.length; i++) {
        dp[i] = new Array(K + 1);
    }
  
    let sum = 1;
    for(i = 1; i <= N; i++)
    {
         
        // dp[i][0] = (dp[i-1][0]+dp[i-1][1]..dp[i-1][k])*21
        dp[i][0] = sum * 21;
        dp[i][0] %= MOD;
  
        // Now setting sum to be dp[i][0]
        sum = dp[i][0];
  
        for(j = 1; j <= K; j++)
        {
             
            // If j>i, no ways are possible to create
            // a string with length i and vowel j
            if (j > i)
                dp[i][j] = 0;
                 
            else if (j == i)
            {
                 
                // If j = i all the character should
                // be vowel
                dp[i][j] = power(5, i, MOD);
            }
            else
            {
                 
                // dp[i][j] relation with dp[i-1][j-1]
                dp[i][j] = dp[i - 1][j - 1] * 5;
            }
  
            dp[i][j] %= MOD;
  
            // Adding dp[i][j] in the sum
            sum += dp[i][j];
            sum %= MOD;
        }
    }
    return sum;
}
 
// Driver Code
 
    // Input
    let N = 3;
    let K = 3;
  
    // Function Call
    document.write( kvowelwords(N, K));
     
    // This code is contributed by sanjoy_62.

输出：

17576

时间复杂度： O(N×K)
辅助空间： O(N×K)

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