给定三个正整数N ， A和B。任务是计算仅包含数字A和B且其数字总和也仅包含数字A和B的长度N的数量。以10 9 + 7的模数打印答案。
例子：
Input: N = 3, A = 1, B = 3
Output: 1
Possible numbers of length 3 are 113, 131, 111, 333, 311, 331 and so on…
But only 111 is a valid number since its sum of digits is 3 (contains digits A and B only)
Input: N = 10, A = 2, B = 3
Output: 165
方法：这个想法是将数字的总和表示为线性方程式，包含两个变量，即
S = X * A + Y * B其中A和B是给定的数字， X和Y分别是这些数字的频率。
由于(X + Y)的总和根据给定条件应等于N (数字的长度)，因此我们可以将Y替换为(N – X) ，等式简化为S = X * A +(N – X)* B。因此，X =(S – N * B)/(A – B)。
现在，我们可以遍历S的所有可能的值，其中S的最小值为N位的数字，其中所有数字是1和S的最大值是N位的数字，其中所有数字都是9，检查如果当前值包含仅数字A和B。使用上述公式为有效电流S求X和Y的值。既然如此，我们还可以将当前值S的数字位数替换为(N！/ X！Y！) 。将此结果添加到最终答案中。
注意：使用费马小定理来计算n！ ％p 。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
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;
} 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 mitsPython3
# 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_GangwarC#
// 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输出：
1