Dados tres enteros positivos N , A y B . La tarea es contar los números de longitud N que contienen solo los dígitos A y B y cuya suma de dígitos también contiene solo los dígitos A y B. Imprime la respuesta módulo 10 9 + 7.
Ejemplos:
Entrada: N = 3, A = 1, B = 3
Salida: 1
Los posibles números de longitud 3 son 113, 131, 111, 333, 311, 331 y así sucesivamente…
Pero solo 111 es un número válido ya que su suma de dígitos es 3 (contiene solo los dígitos A y B)
Entrada: N = 10, A = 2, B = 3
Salida: 165
Enfoque: La idea es expresar la suma de los dígitos del número como una ecuación lineal en dos variables, es decir,
S = X * A + Y * B donde A y B son los dígitos dados y X e Y son las frecuencias de estos dígitos respectivamente . .
Dado que la suma de (X + Y) debe ser igual a N (longitud del número) de acuerdo con la condición dada, podemos reemplazar Y con (N – X) y la ecuación se reduce a S = X * A + (N – X) * B . Así, X = (S – N * B) / (A – B).
Ahora, podemos iterar sobre todos los valores posibles de S donde el valor mínimo de S es un dígito Nnúmero donde todos los dígitos son 1 y el valor máximo de S es un número de N dígitos donde todos los dígitos son 9 y verifique si el valor actual contiene solo los dígitos A y B. Encuentre los valores de X e Y utilizando la fórmula anterior para una S actual válida . Dado que, también podemos permutar el número de dígitos de los números será (N! / X! Y!) para el valor actual S. Agregue este resultado a la respuesta final.
Nota: ¡ Use el pequeño teorema de Fermat para calcular n! % pág .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e5 + 5;
const int MOD = 1e9 + 7;
#define ll long long
// Function that returns true if the num contains
// a and b digits only
int check(int num, int a, int b)
{
while (num) {
int rem = num % 10;
num /= 10;
if (rem != a && rem != b)
return 0;
}
return 1;
}
// Modular Exponentiation
ll power(ll x, ll y)
{
ll ans = 1;
while (y) {
if (y & 1)
ans = (ans * x) % MOD;
y >>= 1;
x = (x * x) % MOD;
}
return ans % MOD;
}
// Function to return the modular inverse
// of x modulo MOD
int modInverse(int x)
{
return power(x, MOD - 2);
}
// Function to return the required count
// of numbers
ll countNumbers(int n, int a, int b)
{
ll fact[MAX], inv[MAX];
ll ans = 0;
// Generating factorials of all numbers
fact[0] = 1;
for (int i = 1; i < MAX; i++) {
fact[i] = (1LL * fact[i - 1] * i);
fact[i] %= MOD;
}
// Generating inverse of factorials modulo
// MOD of all numbers
inv[MAX - 1] = modInverse(fact[MAX - 1]);
for (int i = MAX - 2; i >= 0; i--) {
inv[i] = (inv[i + 1] * (i + 1));
inv[i] %= MOD;
}
// Keeping a as largest number
if (a < b)
swap(a, b);
// Iterate over all possible values of s and
// if it is a valid S then proceed further
for (int s = n; s <= 9 * n; s++) {
if (!check(s, a, b))
continue;
// Check for invalid cases in the equation
if (s < n * b || (s - n * b) % (a - b) != 0)
continue;
int numDig = (s - n * b) / (a - b);
if (numDig > n)
continue;
// Find answer using combinatorics
ll curr = fact[n];
curr = (curr * inv[numDig]) % MOD;
curr = (curr * inv[n - numDig]) % MOD;
// Add this result to final answer
ans = (ans + curr) % MOD;
}
return ans;
}
// Driver Code
int main()
{
int n = 3, a = 1, b = 3;
cout << countNumbers(n, a, b);
return 0;
}
C
// C implementation of the approach
#include <stdio.h>
#include<math.h>
const int MAX = 1e5 + 5;
const int MOD = 1e9 + 7;
#define ll long long
// Function that returns true if the num contains
// a and b digits only
int check(int num, int a, int b)
{
while (num) {
int rem = num % 10;
num /= 10;
if (rem != a && rem != b)
return 0;
}
return 1;
}
// Modular Exponentiation
ll power(ll x, ll y)
{
ll ans = 1;
while (y) {
if (y & 1)
ans = (ans * x) % MOD;
y >>= 1;
x = (x * x) % MOD;
}
return ans % MOD;
}
// Function to return the modular inverse
// of x modulo MOD
int modInverse(int x)
{
return power(x, MOD - 2);
}
// Function to return the required count
// of numbers
ll countNumbers(int n, int a, int b)
{
ll fact[MAX], inv[MAX];
ll ans = 0;
// Generating factorials of all numbers
fact[0] = 1;
for (int i = 1; i < MAX; i++) {
fact[i] = (1LL * fact[i - 1] * i);
fact[i] %= MOD;
}
// Generating inverse of factorials modulo
// MOD of all numbers
inv[MAX - 1] = modInverse(fact[MAX - 1]);
for (int i = MAX - 2; i >= 0; i--) {
inv[i] = (inv[i + 1] * (i + 1));
inv[i] %= MOD;
}
// Keeping a as largest number
if (a < b){
//Swapping a and b
a=a+b;
b=a-b;
a=a-b;
}
// Iterate over all possible values of s and
// if it is a valid S then proceed further
for (int s = n; s <= 9 * n; s++) {
if (!check(s, a, b))
continue;
// Check for invalid cases in the equation
if (s < n * b || (s - n * b) % (a - b) != 0)
continue;
int numDig = (s - n * b) / (a - b);
if (numDig > n)
continue;
// Find answer using combinatorics
ll curr = fact[n];
curr = (curr * inv[numDig]) % MOD;
curr = (curr * inv[n - numDig]) % MOD;
// Add this result to final answer
ans = (ans + curr) % MOD;
}
return ans;
}
// Driver Code
int main()
{
int n = 3, a = 1, b = 3;
printf("%lld",countNumbers(n, a, b));
return 0;
}
// This code is contributed by allwink45.
Java
// Java implementation of the approach
class GFG
{
static int MAX = (int)(1E5 + 5);
static long MOD = (long)(1E9 + 7);
// Function that returns true if the num contains
// a and b digits only
static boolean check(long num, long a, long b)
{
while (num > 0)
{
long rem = num % 10;
num /= 10;
if (rem != a && rem != b)
return false;
}
return true;
}
// Modular Exponentiation
static long power(long x, long y)
{
long ans = 1;
while (y > 0)
{
if ((y & 1) > 0)
ans = (ans * x) % MOD;
y >>= 1;
x = (x * x) % MOD;
}
return ans % MOD;
}
// Function to return the modular inverse
// of x modulo MOD
static long modInverse(long x)
{
return power(x, MOD - 2);
}
// Function to return the required count
// of numbers
static long countNumbers(long n, long a, long b)
{
long[] fact = new long[MAX];
long[] inv = new long[MAX];
long ans = 0;
// Generating factorials of all numbers
fact[0] = 1;
for (int i = 1; i < MAX; i++)
{
fact[i] = (1 * fact[i - 1] * i);
fact[i] %= MOD;
}
// Generating inverse of factorials modulo
// MOD of all numbers
inv[MAX - 1] = modInverse(fact[MAX - 1]);
for (int i = MAX - 2; i >= 0; i--)
{
inv[i] = (inv[i + 1] * (i + 1));
inv[i] %= MOD;
}
// Keeping a as largest number
if (a < b)
{
long x = a;
a = b;
b = x;
}
// Iterate over all possible values of s and
// if it is a valid S then proceed further
for (long s = n; s <= 9 * n; s++)
{
if (!check(s, a, b))
continue;
// Check for invalid cases in the equation
if (s < n * b || (s - n * b) % (a - b) != 0)
continue;
int numDig = (int)((s - n * b) / (a - b));
if (numDig > n)
continue;
// Find answer using combinatorics
long curr = fact[(int)n];
curr = (curr * inv[numDig]) % MOD;
curr = (curr * inv[(int)n - numDig]) % MOD;
// Add this result to final answer
ans = (ans + curr) % MOD;
}
return ans;
}
// Driver Code
public static void main (String[] args)
{
long n = 3, a = 1, b = 3;
System.out.println(countNumbers(n, a, b));
}
}
// This code is contributed by mits
Python3
# Python 3 implementation of the approach MAX = 100005; MOD = 1000000007 # Function that returns true if the num # contains a and b digits only def check(num, a, b): while (num): rem = num % 10 num = int(num / 10) if (rem != a and rem != b): return 0 return 1 # Modular Exponentiation def power(x, y): ans = 1 while (y): if (y & 1): ans = (ans * x) % MOD y >>= 1 x = (x * x) % MOD return ans % MOD # Function to return the modular # inverse of x modulo MOD def modInverse(x): return power(x, MOD - 2) # Function to return the required # count of numbers def countNumbers(n, a, b): fact = [0 for i in range(MAX)] inv = [0 for i in range(MAX)] ans = 0 # Generating factorials of all numbers fact[0] = 1 for i in range(1, MAX, 1): fact[i] = (1 * fact[i - 1] * i) fact[i] %= MOD # Generating inverse of factorials # modulo MOD of all numbers inv[MAX - 1] = modInverse(fact[MAX - 1]) i = MAX - 2 while(i >= 0): inv[i] = (inv[i + 1] * (i + 1)) inv[i] %= MOD i -= 1 # Keeping a as largest number if (a < b): temp = a a = b b = temp # Iterate over all possible values of s and # if it is a valid S then proceed further for s in range(n, 9 * n + 1, 1): if (check(s, a, b) == 0): continue # Check for invalid cases in the equation if (s < n * b or (s - n * b) % (a - b) != 0): continue numDig = int((s - n * b) / (a - b)) if (numDig > n): continue # Find answer using combinatorics curr = fact[n] curr = (curr * inv[numDig]) % MOD curr = (curr * inv[n - numDig]) % MOD # Add this result to final answer ans = (ans + curr) % MOD return ans # Driver Code if __name__ == '__main__': n = 3 a = 1 b = 3 print(countNumbers(n, a, b)) # This code is contributed by # Surendra_Gangwar
C#
// C# implementation of the approach
using System;
class GFG
{
static long MAX = (long)(1E5 + 5);
static long MOD = (long)(1E9 + 7);
// Function that returns true if the num contains
// a and b digits only
static bool check(long num, long a, long b)
{
while (num > 0)
{
long rem = num % 10;
num /= 10;
if (rem != a && rem != b)
return false;
}
return true;
}
// Modular Exponentiation
static long power(long x, long y)
{
long ans = 1;
while (y > 0)
{
if ((y & 1) > 0)
ans = (ans * x) % MOD;
y >>= 1;
x = (x * x) % MOD;
}
return ans % MOD;
}
// Function to return the modular inverse
// of x modulo MOD
static long modInverse(long x)
{
return power(x, MOD - 2);
}
// Function to return the required count
// of numbers
static long countNumbers(long n, long a, long b)
{
long[] fact = new long[MAX];
long[] inv = new long[MAX];
long ans = 0;
// Generating factorials of all numbers
fact[0] = 1;
for (long i = 1; i < MAX; i++)
{
fact[i] = (1 * fact[i - 1] * i);
fact[i] %= MOD;
}
// Generating inverse of factorials modulo
// MOD of all numbers
inv[MAX - 1] = modInverse(fact[MAX - 1]);
for (long i = MAX - 2; i >= 0; i--)
{
inv[i] = (inv[i + 1] * (i + 1));
inv[i] %= MOD;
}
// Keeping a as largest number
if (a < b)
{
long x = a;
a = b;
b = x;
}
// Iterate over all possible values of s and
// if it is a valid S then proceed further
for (long s = n; s <= 9 * n; s++)
{
if (!check(s, a, b))
continue;
// Check for invalid cases in the equation
if (s < n * b || (s - n * b) % (a - b) != 0)
continue;
long numDig = (s - n * b) / (a - b);
if (numDig > n)
continue;
// Find answer using combinatorics
long curr = fact[n];
curr = (curr * inv[numDig]) % MOD;
curr = (curr * inv[n - numDig]) % MOD;
// Add this result to final answer
ans = (ans + curr) % MOD;
}
return ans;
}
// Driver Code
static void Main()
{
long n = 3, a = 1, b = 3;
Console.WriteLine(countNumbers(n, a, b));
}
}
// This code is contributed by mits
Javascript
<script>
// JavaScript implementation of the approach
const MAX = (5);
let MOD = (7);
// Function that returns true if the num contains
// a and b digits only
function check(num , a , b) {
while (num > 0) {
var rem = num % 10;
num = parseInt(num/10);
if (rem != a && rem != b)
return false;
}
return true;
}
// Modular Exponentiation
function power(x , y) {
var ans = 1;
while (y > 0) {
if ((y & 1) > 0)
ans = (ans * x) % MOD;
y >>= 1;
x = (x * x) % MOD;
}
return ans % MOD;
}
// Function to return the modular inverse
// of x modulo MOD
function modInverse(x) {
return power(x, MOD - 2);
}
// Function to return the required count
// of numbers
function countNumbers(n , a , b) {
var fact = Array(MAX).fill(0);
var inv = Array(MAX).fill(0);
var ans = 0;
// Generating factorials of all numbers
fact[0] = 1;
for (var i = 1; i < MAX; i++) {
fact[i] = (1 * fact[i - 1] * i);
fact[i] %= MOD;
}
// Generating inverse of factorials modulo
// MOD of all numbers
inv[MAX - 1] = modInverse(fact[MAX - 1]);
for (i = MAX - 2; i >= 0; i--) {
inv[i] = (inv[i + 1] * (i + 1));
inv[i] %= MOD;
}
// Keeping a as largest number
if (a < b) {
var x = a;
a = b;
b = x;
}
// Iterate over all possible values of s and
// if it is a valid S then proceed further
for (s = n; s <= 9 * n; s++) {
if (!check(s, a, b))
continue;
// Check for invalid cases in the equation
if (s < n * b || (s - n * b) % (a - b) != 0)
continue;
var numDig = parseInt( ((s - n * b) / (a - b)));
if (numDig > n)
continue;
// Find answer using combinatorics
var curr = fact[parseInt(n)];
curr = (curr * inv[numDig]) % MOD;
curr = (curr * inv[parseInt( n - numDig)]) % MOD;
// Add this result to final answer
ans = (ans + curr) % MOD;
}
return ans;
}
// Driver Code
var n = 3, a = 1, b = 3;
document.write(countNumbers(n, a, b));
// This code contributed by umadevi9616
</script>
Producción
1
Complejidad de tiempo: O (máx.)
Espacio Auxiliar: O(Max), ya que se ha tomado Max espacio extra.
Publicación traducida automáticamente
Artículo escrito por Nishant Tanwar y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA