长度为 N 的数字，具有数字 A 和 B，并且其数字总和仅包含数字 A 和 B
原文:https://www . geesforgeks . org/numbers-of-length-n-having-digits-a-b-and-what-sum-of-digits-a-b/
给定三个正整数 N 、 A 和 B 。任务是计算长度为 N 的数字，其中只包含数字 A 和 B，并且其数字总和也只包含数字 A 和 B。打印答案模 10 9 + 7。 举例:
输入: N = 3，A = 1，B = 3 输出: 1 长度为 3 的可能数字有 113、131、111、333、311、331 等等…… 但只有 111 是有效数字，因为它的数字之和是 3(只包含数字 A 和 B) 输入: N = 10，A = 2，B = 3 输出:…
方法:思路是将数字的位数之和表示为两个变量中的线性方程，即 S = X * A + Y * B 其中 A 和 B 为给定的位数， X 和 Y 分别为这些位数的频率。 既然 (X + Y) 的和根据给定的条件应该等于 N (数字的长度)，我们就可以用(N–X)代替 Y ，方程化简为S = X * A+(N–X)* B。因此，X =(S–N * B)/(A–B)。 现在，我们可以迭代所有可能的 S 值，其中 S 的最小值是一个 N 位数字，其中所有数字都是 1 ，S 的最大值是一个 N 位数字，其中所有数字都是 9 ，并检查当前值是否只包含数字 A 和 B 。使用以上有效电流 S 公式求出 X 和 Y 的值。既然如此，我们也可以将数字的位数置换为 (N！/ X！y！)为当前值 s .将此结果添加到最终答案中。 注:用费马小定理计算 n！% p 。 以下是上述方法的实施:
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;
}
Java 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
java 描述语言
<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>
Output
1
时间复杂度: O(最大值)
辅助空间:0(最大)
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