给定大小为N的数组arr [] ,其中每个元素为0或1 。任务是找到在素数索引处的0和1的计数。
例子:
Input: arr[] = {1, 0, 1, 0, 1}
Output:
Number of 0s = 1
Number of 1s = 1
Input: arr[] = {1, 0, 1, 1}
Output:
Number of 0s = 0
Number of 1s = 2
方法:遍历数组,如果当前索引为质数,则遇到的每个0将更新计数0,并为质数索引的所有1更新计数1。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
// Function that returns true
// if n is prime
bool isPrime(int n)
{
if (n <= 1)
return false;
// Check from 2 to n
for (int i = 2; i < n; i++)
{
if (n % i == 0)
return false;
}
return true;
}
// Function to find the count
// of 0s and 1s at prime indices
void countPrimePosition(int arr[], int n)
{
// To store the count of 0s and 1s
int c0 = 0, c1 = 0;
for (int i = 0; i < n; i++)
{
// If current 0 is at
// prime position
if (arr[i] == 0 && isPrime(i))
c0++;
// If current 1 is at
// prime position
if (arr[i] == 1 && isPrime(i))
c1++;
}
cout << "Number of 0s = " << c0 << endl;
cout << "Number of 1s = " << c1;
}
// Driver code
int main()
{
int arr [] = { 1, 0, 1, 0, 1 };
int n = sizeof(arr)/ sizeof(arr[0]);
countPrimePosition(arr, n);
return 0;
}
// This code is contributed by ihritik
Java
// Java implementation of the approach
class GFG {
// Function that returns true
// if n is prime
static boolean isPrime(int n)
{
if (n <= 1)
return false;
// Check from 2 to n
for (int i = 2; i < n; i++) {
if (n % i == 0)
return false;
}
return true;
}
// Function to find the count
// of 0s and 1s at prime indices
static void countPrimePosition(int arr[])
{
// To store the count of 0s and 1s
int c0 = 0, c1 = 0;
int n = arr.length;
for (int i = 0; i < n; i++) {
// If current 0 is at
// prime position
if (arr[i] == 0 && isPrime(i))
c0++;
// If current 1 is at
// prime position
if (arr[i] == 1 && isPrime(i))
c1++;
}
System.out.println("Number of 0s = " + c0);
System.out.println("Number of 1s = " + c1);
}
// Driver code
public static void main(String[] args)
{
int[] arr = { 1, 0, 1, 0, 1 };
countPrimePosition(arr);
}
}
Python3
# Python3 implementation of the approach
# Function that returns true
# if n is prime
def isPrime(n) :
if (n <= 1) :
return False;
# Check from 2 to n
for i in range(2, n) :
if (n % i == 0) :
return False;
return True;
# Function to find the count
# of 0s and 1s at prime indices
def countPrimePosition(arr) :
# To store the count of 0s and 1s
c0 = 0; c1 = 0;
n = len(arr);
for i in range(n) :
# If current 0 is at
# prime position
if (arr[i] == 0 and isPrime(i)) :
c0 += 1;
# If current 1 is at
# prime position
if (arr[i] == 1 and isPrime(i)) :
c1 += 1;
print("Number of 0s =", c0);
print("Number of 1s =", c1);
# Driver code
if __name__ == "__main__" :
arr = [ 1, 0, 1, 0, 1 ];
countPrimePosition(arr);
# This code is contributed by AnkitRai01
C#
// C# implementation of the approach
using System;
class GFG
{
// Function that returns true
// if n is prime
static bool isPrime(int n)
{
if (n <= 1)
return false;
// Check from 2 to n
for (int i = 2; i < n; i++)
{
if ((n % i) == 0)
return false;
}
return true;
}
// Function to find the count
// of 0s and 1s at prime indices
static void countPrimePosition(int []arr)
{
// To store the count of 0s and 1s
int c0 = 0, c1 = 0;
int n = arr.Length;
for (int i = 0; i < n; i++)
{
// If current 0 is at
// prime position
if ((arr[i] == 0) && (isPrime(i)))
c0++;
// If current 1 is at
// prime position
if ((arr[i] == 1) && (isPrime(i)))
c1++;
}
Console.WriteLine("Number of 0s = " + c0);
Console.WriteLine("Number of 1s = " + c1);
}
// Driver code
static public void Main ()
{
int[] arr = { 1, 0, 1, 0, 1 };
countPrimePosition(arr);
}
}
// This code is contributed by ajit.
输出:
Number of 0s = 1
Number of 1s = 1