计数长度为 N 且只有 0 和 1 的二进制字符串的数量
原文:https://www . geesforgeks . org/count-number-of-binary-strings-of-length-n-having-only-0s-and-1s/
给定一个整数 N ，任务是计算长度为 n 的二进制字符串的数量，这些字符串只有 0 和 1。注意: T3】由于计数可能非常大，所以返回模 10^9+7.的答案
示例:
输入: 2 输出: 4 说明:数字为 00、01、11、10。因此计数为 4。
输入: 3 输出: 8 说明:数字为 000、001、011、010、111、101、110、100。因此计数是 8。
方法:使用排列组合可以很容易地解决问题。在字符串的每个位置只能有两种可能性，即 0 或 1。因此，长度为 n 的字符串中 0 和 1 的置换总数由 2 * 2 * 2 *……(n 次)给出，即 2^N.的答案可能非常大，因此返回 10^9+7 的模。
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
#define mod (ll)(1e9 + 7)
// Iterative Function to calculate (x^y)%p in O(log y)
ll power(ll x, ll y, ll p)
{
ll res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0) {
// If y is odd, multiply x with result
if (y & 1)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
// Function to count the number of binary
// strings of length N having only 0's and 1's
ll findCount(ll N)
{
int count = power(2, N, mod);
return count;
}
// Driver code
int main()
{
ll N = 25;
cout << findCount(N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the above approach
import java.util.*;
class GFG
{
static int mod = (int) (1e9 + 7);
// Iterative Function to calculate (x^y)%p in O(log y)
static int power(int x, int y, int p)
{
int res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with result
if ((y & 1)==1)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
// Function to count the number of binary
// strings of length N having only 0's and 1's
static int findCount(int N)
{
int count = power(2, N, mod);
return count;
}
// Driver code
public static void main(String[] args)
{
int N = 25;
System.out.println(findCount(N));
}
}
/* This code contributed by PrinciRaj1992 */
Python 3
# Python 3 implementation of the approach
mod = 1000000007
# Iterative Function to calculate (x^y)%p in O(log y)
def power(x, y, p):
res = 1 # Initialize result
x = x % p # Update x if it is more than or
# equal to p
while (y > 0):
# If y is odd, multiply x with result
if (y & 1):
res = (res * x) % p
# y must be even now
y = y >> 1 # y = y/2
x = (x * x) % p
return res
# Function to count the number of binary
# strings of length N having only 0's and 1's
def findCount(N):
count = power(2, N, mod)
return count
# Driver code
if __name__ == '__main__':
N = 25
print(findCount(N))
# This code is contributed by
# Surendra_Gangwar
C
// C# implementation of the above approach
using System;
class GFG
{
static int mod = (int) (1e9 + 7);
// Iterative Function to calculate (x^y)%p in O(log y)
static int power(int x, int y, int p)
{
int res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with result
if ((y & 1) == 1)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
// Function to count the number of binary
// strings of length N having only 0's and 1's
static int findCount(int N)
{
int count = power(2, N, mod);
return count;
}
// Driver code
public static void Main()
{
int N = 25;
Console.WriteLine(findCount(N));
}
}
// This code is contributed by Ryuga
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP implementation of the approach
// Iterative Function to calculate
// (x^y)%p in O(log y)
function power($x, $y)
{
$p = 1000000007;
$res = 1; // Initialize result
$x = $x % $p; // Update x if it is more
// than or equal to p
while ($y > 0)
{
// If y is odd, multiply x with result
if ($y & 1)
$res = ($res * $x) % $p;
// y must be even now
$y = $y >> 1; // y = y/2
$x = ($x * $x) % $p;
}
return $res;
}
// Function to count the number of binary
// strings of length N having only 0's and 1's
function findCount($N)
{
$count = power(2, $N);
return $count;
}
// Driver code
$N = 25;
echo findCount($N);
// This code is contributed by Rajput-Ji
?>
java 描述语言
<script>
// Javascript implementation of the approach
mod = 1000000007
// Iterative Function to calculate
// (x^y)%p in O(log y)
function power(x, y, p)
{
// Initialize result
var res = 1;
// Update x if it is more than or
// equal to p
x = x % p;
while (y > 0)
{
// If y is odd, multiply x with result
if (y & 1)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
// Function to count the number of binary
// strings of length N having only 0's and 1's
function findCount(N)
{
var count = power(2, N, mod);
return count;
}
// Driver code
var N = 25;
document.write(findCount(N));
// This code is contributed by noob2000
</script>
Output:
33554432
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