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Posekių skaičius eilutėje, kuri dalijasi iš n

Duota eilutė, susidedanti iš skaitmenų nuo 0 iki 9, suskaičiuokite joje esančių posekių skaičių, dalijamą iš m.
Pavyzdžiai:  
 

Input : str = '1234' n = 4  
Output : 4
The subsequences 4 12 24 and 124 are
divisible by 4.

Input : str = '330' n = 6
Output : 4
The subsequences 30 30 330 and 0 are
divisible by n.
Input : str = '676' n = 6
Output : 3
The subsequences 6 6 and 66


 

Rekomenduojama praktika Posekių skaičius eilutėje, kuri dalijasi iš nIšbandyk!


Šią problemą galima apibrėžti rekursyviai. Tegul likusi eilutės, kurios reikšmė x, dalis yra „r“, padalyta iš n. Pridėjus dar vieną simbolį prie šios eilutės, jos likutis pakeičiamas į (r*10 + naujas skaitmuo) % n. Kiekvienam naujam simboliui turime du pasirinkimus arba pridėti jį prie visų dabartinių posekių arba ignoruoti. Taigi turime optimalią pagrindą. Toliau parodyta brutalios jėgos versija:
 



string str = '330';  
int n = 6
// idx is value of current index in str
// rem is current remainder
int count(int idx int rem)
{
// If last character reached
if (idx == n)
return (rem == 0)? 1 : 0;
int ans = 0;
// we exclude it thus remainder
// remains the same
ans += count(idx+1 rem);
// we include it and thus new remainder
ans += count(idx+1 (rem*10 + str[idx]-'0')%n);
return ans;
}


Aukščiau pateiktas rekursinis sprendimas turi sutampančių subproblemų, kaip parodyta žemiau esančiame rekursijos medyje.
 

 input string = '330'  
(00) ===> at 0th index with 0 remainder
(exclude 0th / (include 0th character)
character) /
(10) (13) ======> at index 1 with 3 as
(E)/ (I) /(E) the current remainder
(20) (23) (23)
|-------|
These two subproblems overlap


Taigi galime pritaikyti dinaminį programavimą. Žemiau pateikiamas įgyvendinimas.
 

C++
// C++ program to count subsequences of a // string divisible by n. #include   using namespace std; // Returns count of subsequences of str // divisible by n. int countDivisibleSubseq(string str int n) {  int len = str.length();  // division by n can leave only n remainder  // [0..n-1]. dp[i][j] indicates number of  // subsequences in string [0..i] which leaves  // remainder j after division by n.  int dp[len][n];  memset(dp 0 sizeof(dp));  // Filling value for first digit in str  dp[0][(str[0]-'0')%n]++;  for (int i=1; i<len; i++)  {  // start a new subsequence with index i  dp[i][(str[i]-'0')%n]++;  for (int j=0; j<n; j++)  {  // exclude i'th character from all the  // current subsequences of string [0...i-1]  dp[i][j] += dp[i-1][j];  // include i'th character in all the current  // subsequences of string [0...i-1]  dp[i][(j*10 + (str[i]-'0'))%n] += dp[i-1][j];  }  }  return dp[len-1][0]; } // Driver code int main() {  string str = '1234';  int n = 4;  cout << countDivisibleSubseq(str n);  return 0; } 
Java
//Java program to count subsequences of a // string divisible by n class GFG { // Returns count of subsequences of str // divisible by n.  static int countDivisibleSubseq(String str int n) {  int len = str.length();  // division by n can leave only n remainder  // [0..n-1]. dp[i][j] indicates number of  // subsequences in string [0..i] which leaves  // remainder j after division by n.  int dp[][] = new int[len][n];  // Filling value for first digit in str  dp[0][(str.charAt(0) - '0') % n]++;  for (int i = 1; i < len; i++) {  // start a new subsequence with index i  dp[i][(str.charAt(i) - '0') % n]++;  for (int j = 0; j < n; j++) {  // exclude i'th character from all the  // current subsequences of string [0...i-1]  dp[i][j] += dp[i - 1][j];  // include i'th character in all the current  // subsequences of string [0...i-1]  dp[i][(j * 10 + (str.charAt(i) - '0')) % n] += dp[i - 1][j];  }  }  return dp[len - 1][0];  } // Driver code  public static void main(String[] args) {  String str = '1234';  int n = 4;  System.out.print(countDivisibleSubseq(str n));  } } // This code is contributed by 29AjayKumar 
Python 3
# Python 3 program to count subsequences  # of a string divisible by n. # Returns count of subsequences of  # str divisible by n. def countDivisibleSubseq(str n): l = len(str) # division by n can leave only n remainder # [0..n-1]. dp[i][j] indicates number of # subsequences in string [0..i] which leaves # remainder j after division by n. dp = [[0 for x in range(l)] for y in range(n)] # Filling value for first digit in str dp[int(str[0]) % n][0] += 1 for i in range(1 l): # start a new subsequence with index i dp[int(str[i]) % n][i] += 1 for j in range(n): # exclude i'th character from all the # current subsequences of string [0...i-1] dp[j][i] += dp[j][i-1] # include i'th character in all the current # subsequences of string [0...i-1] dp[(j * 10 + int(str[i])) % n][i] += dp[j][i-1] return dp[0][l-1] # Driver code if __name__ == '__main__': str = '1234' n = 4 print(countDivisibleSubseq(str n)) # This code is contributed by ita_c 
C#
//C# program to count subsequences of a // string divisible by n   using System; class GFG {   // Returns count of subsequences of str // divisible by n.  static int countDivisibleSubseq(string str int n) {  int len = str.Length;    // division by n can leave only n remainder  // [0..n-1]. dp[i][j] indicates number of  // subsequences in string [0..i] which leaves  // remainder j after division by n.  int[] dp = new int[lenn];    // Filling value for first digit in str  dp[0(str[0] - '0') % n]++;    for (int i = 1; i < len; i++) {  // start a new subsequence with index i  dp[i(str[i] - '0') % n]++;    for (int j = 0; j < n; j++) {  // exclude i'th character from all the  // current subsequences of string [0...i-1]  dp[ij] += dp[i - 1j];    // include i'th character in all the current  // subsequences of string [0...i-1]  dp[i(j * 10 + (str[i] - '0')) % n] += dp[i - 1j];  }  }    return dp[len - 10];  }   // Driver code  public static void Main() {  String str = '1234';  int n = 4;  Console.Write(countDivisibleSubseq(str n));  } } 
JavaScript
<script> //Javascript program to count subsequences of a // string divisible by n    // Returns count of subsequences of str  // divisible by n.  function countDivisibleSubseq(strn)  {  let len = str.length;    // division by n can leave only n remainder  // [0..n-1]. dp[i][j] indicates number of  // subsequences in string [0..i] which leaves  // remainder j after division by n.  let dp = new Array(len);  for(let i = 0; i < len; i++)  {  dp[i] = new Array(n);  for(let j = 0; j < n; j++)  {  dp[i][j] = 0;  }  }    // Filling value for first digit in str  dp[0][(str[0] - '0') % n]++;    for (let i = 1; i < len; i++) {  // start a new subsequence with index i  dp[i][(str[i] - '0') % n]++;    for (let j = 0; j < n; j++) {  // exclude i'th character from all the  // current subsequences of string [0...i-1]  dp[i][j] += dp[i - 1][j];    // include i'th character in all the current  // subsequences of string [0...i-1]  dp[i][(j * 10 + (str[i] - '0')) % n] += dp[i - 1][j];  }  }    return dp[len - 1][0];  }    // Driver code  let str = '1234';  let n = 4;  document.write(countDivisibleSubseq(str n));    // This code is contributed by avanitrachhadiya2155 </script>  

Išvestis
4

Laiko sudėtingumas: O (len * n) 
Pagalbinė erdvė: O (len * n)


Efektyvus metodas: erdvės optimizavimas

Ankstesniame požiūryje dp[i][j] priklauso nuo dabartinės ir ankstesnės 2D matricos eilutės. Taigi, norėdami optimizuoti erdvę, naudojame du vektorius temp ir dp kurie seka dabartinę ir ankstesnę eilutę DP .

Diegimo žingsniai:

  • The countDivisibleSubseq funkcija skaičiuoja tam tikros eilutės str posekių, kurios dalijasi iš nurodyto skaičiaus n, skaičių.
  • Jis inicijuoja masyvą dp dydžio n skaičiams saugoti.
  • Jis kartoja kiekvieną eilutės skaitmenį ir atnaujina skaičių dp remiantis likučiais.
  • Kiekvienoje iteracijoje jis atsižvelgia į dabartinį skaitmenį ir ankstesnius skaičius, kad apskaičiuotų atnaujintus skaičius.
  • Galiausiai jis grąžina saugomų posekių skaičių, dalijamą iš n dp[0] .

Įgyvendinimas:

C++
#include   using namespace std; int countDivisibleSubseq(string str int n) {  int len = str.length();  int dp[n];  memset(dp 0 sizeof(dp));  dp[(str[0]-'0')%n]++; // Increment the count of remainder of first digit by n  for (int i=1; i<len; i++)  {  int temp[n];  memset(temp 0 sizeof(temp));  temp[(str[i]-'0')%n]++; // Increment the count of remainder of current digit by n  for (int j=0; j<n; j++)  {  temp[j] += dp[j]; // Carry over the counts from previous digit  temp[(j*10 + (str[i]-'0'))%n] += dp[j]; // Update the count with the new remainder formed by appending the current digit  }  for (int j=0; j<n; j++)  {  dp[j] = temp[j]; // Copy the updated counts from temp back to dp for the next iteration  }  }  return dp[0]; // Return the count of subsequences divisible by n } int main() {  string str = '1234';  int n = 4;  cout << countDivisibleSubseq(str n);  return 0; } 
Java
// Java program to count subsequences // of a string divisible by n. public class GFG {  public static int countDivisibleSubseq(String str int n) {  int length = str.length();  int[] dp = new int[n]; // Create an array of size n to store counts  // Increment the count of remainder of first digit by n  dp[Integer.parseInt(String.valueOf(str.charAt(0))) % n] += 1;  for (int i = 1; i < length; i++) {  int[] temp = new int[n]; // Create a temporary array of size n  // Increment the count of remainder of current digit by n  temp[Integer.parseInt(String.valueOf(str.charAt(i))) % n] += 1;  for (int j = 0; j < n; j++) {  temp[j] += dp[j]; // Carry over the counts from the previous digit  // Calculate the new remainder  int newRemainder = (j * 10 + Integer.parseInt(String.valueOf(str.charAt(i)))) % n;  // Update the count with the new remainder  temp[newRemainder] += dp[j];  }  dp = temp;  }  return dp[0];  } //Driver code  public static void main(String[] args) {  String str = '1234';  int n = 4;  System.out.println(countDivisibleSubseq(str n));  } } 
Python3
# Python 3 program to count subsequences # of a string divisible by n. def countDivisibleSubseq(str n): length = len(str) dp = [0] * n # Create an array of size n # Increment the count of remainder of first digit by n dp[int(str[0]) % n] += 1 for i in range(1 length): temp = [0] * n # Create a temporary array of size n # Increment the count of remainder of current digit by n temp[int(str[i]) % n] += 1 for j in range(n): temp[j] += dp[j] # Carry over the counts from the previous digit # Calculate the new remainder new_remainder = (j * 10 + int(str[i])) % n # Update the count with the new remainder temp[new_remainder] += dp[j] dp = temp return dp[0] # Driver code str = '1234' n = 4 print(countDivisibleSubseq(str n)) 
C#
using System; class GFG {  static int CountDivisibleSubseq(string str int n)  {  int len = str.Length;  int[] dp = new int[n];  Array.Fill(dp 0);  dp[(str[0] - '0')  % n]++; // Increment the count of remainder of  // first digit by n  for (int i = 1; i < len; i++) {  int[] temp = new int[n];  Array.Fill(temp 0);  temp[(str[i] - '0')  % n]++; // Increment the count of remainder  // of current digit by n  for (int j = 0; j < n; j++) {  temp[j] += dp[j]; // Carry over the counts  // from previous digit  temp[(j * 10 + (str[i] - '0')) % n]  += dp[j]; // Update the count with the  // new remainder formed by  // appending the current digit  }  for (int j = 0; j < n; j++) {  dp[j] = temp[j]; // Copy the updated counts  // from temp back to dp for  // the next iteration  }  }  return dp[0]; // Return the count of subsequences  // divisible by n  }  static void Main()  {  string str = '1234';  int n = 4;  Console.WriteLine(CountDivisibleSubseq(str n));  } } 
JavaScript
function countDivisibleSubseq(str n) {  const len = str.length;  const dp = new Array(n).fill(0);    // Increment the count of remainder of first digit by n  dp[Number(str[0]) % n]++;   for (let i = 1; i < len; i++) {  const temp = new Array(n).fill(0);    // Increment the count of remainder of current digit by n  temp[Number(str[i]) % n]++;  for (let j = 0; j < n; j++) {  temp[j] += dp[j]; // Carry over the counts from previous digit  // Update the count with the new remainder   // formed by appending the current digit  temp[(j * 10 + Number(str[i])) % n] += dp[j];   }  for (let j = 0; j < n; j++) {  // Copy the updated counts from   // temp back to dp for the next iteration  dp[j] = temp[j];   }  }  return dp[0]; // Return the count of subsequences divisible by n } const str = '1234'; const n = 4; console.log(countDivisibleSubseq(str n)); 

Išvestis
4

Laiko sudėtingumas: O (len * n) 
Pagalbinė erdvė: O(n)




 

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