计算没有前导零的 N 0 和 M 1 的数字
原文:https://www . geesforgeks . org/count-numbers-having-n-0s-and-m-1s-不带前导零/
给定两个整数 N 和 M ，任务是找出不含前导零的 N 0 和 M 1 以及 N + M 总位数的不同数字。
示例:
输入: N = 2，M = 2 输出: 3 数字为 1001、1010、1100。
输入: N = 2，M = 3 输出: 6 数字为 10011、10101、10110、11001、11010、11100。
方法:通过求 N 相似项和M–1相似项的全排列，可以很容易地解决问题。由于不允许前导零，11始终固定在数字的开头。剩余的M–11 和 N 0 以不同的排列方式排列。现在 (A + B) 对象的排列数，其中 A 同类型对象和 B 其他类型对象由 (A + B)给出！/(一！* B！)。因此，相异数的个数由 (N + M -1)给出！/ (N！*(M–1)！)。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
// Function to return the factorial of a number
ll factorial(int f)
{
ll fact = 1;
for (int i = 2; i <= f; i++)
fact *= (ll)i;
return fact;
}
// Function to return the count of distinct
// (N + M) digit numbers having N 0's
// and and M 1's with no leading zeros
ll findPermutation(int N, int M)
{
ll permutation = factorial(N + M - 1)
/ (factorial(N) * factorial(M - 1));
return permutation;
}
// Driver code
int main()
{
int N = 3, M = 3;
cout << findPermutation(N, M);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
import java.io.*;
class GFG
{
// Function to return the factorial of a number
static int factorial(int f)
{
int fact = 1;
for (int i = 2; i <= f; i++)
fact *= (int)i;
return fact;
}
// Function to return the count of distinct
// (N + M) digit numbers having N 0's
// and and M 1's with no leading zeros
static int findPermutation(int N, int M)
{
int permutation = factorial(N + M - 1)
/ (factorial(N) * factorial(M - 1));
return permutation;
}
// Driver code
public static void main (String[] args)
{
int N = 3, M = 3;
System.out.println(findPermutation(N, M));
}
}
// This code is contributed
// by ajit
Python 3
# Python3 implementation of the approach
# Function to return the factorial
# of a number
def factorial(f):
fact = 1
for i in range(2, f + 1):
fact *= i
return fact
# Function to return the count of distinct
# (N + M) digit numbers having N 0's
# and and M 1's with no leading zeros
def findPermuatation(N, M):
permutation = (factorial(N + M - 1) //
(factorial(N) * factorial(M - 1)))
return permutation
# Driver code
N = 3; M = 3
print(findPermuatation(N, M))
# This code is contributed
# by Shrikant13
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the factorial of a number
static int factorial(int f)
{
int fact = 1;
for (int i = 2; i <= f; i++)
fact *= (int)i;
return fact;
}
// Function to return the count of distinct
// (N + M) digit numbers having N 0's
// and and M 1's with no leading zeros
static int findPermutation(int N, int M)
{
int permutation = factorial(N + M - 1)
/ (factorial(N) * factorial(M - 1));
return permutation;
}
// Driver code
public static void Main()
{
int N = 3, M = 3;
Console.Write(findPermutation(N, M));
}
}
// This code is contributed
// by Akanksha Rai
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Function to return the factorial of a number
function factorial($f)
{
$fact = 1;
for ($i = 2; $i <= $f; $i++)
$fact *= $i;
return $fact;
}
// Function to return the count of distinct
// (N + M) digit numbers having N 0's
// and and M 1's with no leading zeros
function findPermutation($N,$M)
{
$permutation = factorial($N + $M - 1)
/ (factorial($N) * factorial($M - 1));
return $permutation;
}
// Driver code
$N = 3;
$M = 3;
echo findPermutation($N, $M);
// This code is contributed by ajit..
?>
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the factorial of a number
function factorial(f)
{
var fact = 1;
for(var i = 2; i <= f; i++)
fact *= i;
return fact;
}
// Function to return the count of distinct
// (N + M) digit numbers having N 0's
// and and M 1's with no leading zeros
function findPermutation(N, M)
{
var permutation = factorial(N + M - 1) /
(factorial(N) * factorial(M - 1));
return permutation;
}
// Driver code
var N = 3, M = 3;
document.write(findPermutation(N, M));
// This code is contributed by noob2000
</script>
Output:
10
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