给定一个正整数,检查该数字是否为素数。素数是一个大于1的自然数,除1和它本身外,没有其他正因子。前几个素数的例子是{2,3,5, 例如:
null
Input: n = 11Output: trueInput: n = 15Output: falseInput: n = 1Output: false
学校方法 一个简单的解决方案是迭代从2到n-1的所有数字,并对每个数字检查它是否除以n。如果我们找到任何除以的数字,我们将返回false。 下面是这个方法的实现。
C++
// A school method based C++ program to check if a // number is prime #include <bits/stdc++.h> using namespace std; bool isPrime( int n) { // Corner case if (n <= 1) return false ; // Check from 2 to n-1 for ( int i=2; i<n; i++) if (n%i == 0) return false ; return true ; } // Driver Program to test above function int main() { isPrime(11)? cout << " true" : cout << " false" ; isPrime(15)? cout << " true" : cout << " false" ; return 0; } |
JAVA
// A school method based JAVA program // to check if a number is prime class GFG { static boolean isPrime( int n) { // Corner case if (n <= 1 ) return false ; // Check from 2 to n-1 for ( int i = 2 ; i < n; i++) if (n % i == 0 ) return false ; return true ; } // Driver Program public static void main(String args[]) { if (isPrime( 11 )) System.out.println( " true" ); else System.out.println( " false" ); if (isPrime( 15 )) System.out.println( " true" ); else System.out.println( " false" ); } } // This code is contributed // by Nikita Tiwari. |
Python3
# A school method based Python3 # program to check if a number # is prime def isPrime(n): # Corner case if n < = 1 : return False # Check from 2 to n-1 for i in range ( 2 , n): if n % i = = 0 : return False ; return True # Driver Program to test above function print ( "true" ) if isPrime( 11 ) else print ( "false" ) print ( "true" ) if isPrime( 14 ) else print ( "false" ) # This code is contributed by Smitha Dinesh Semwal |
C#
// A optimized school method based C# // program to check if a number is prime using System; namespace prime { public class GFG { public static bool isprime( int n) { // Corner cases if (n <= 1) return false ; for ( int i = 2; i < n; i++) if (n % i == 0) return false ; return true ; } // Driver program public static void Main() { if (isprime(11)) Console.WriteLine( "true" ); else Console.WriteLine( "false" ); if (isprime(15)) Console.WriteLine( "true" ); else Console.WriteLine( "false" ); } } } // This code is contributed by Sam007 |
PHP
<?php // A school method based PHP // program to check if a number // is prime function isPrime( $n ) { // Corner case if ( $n <= 1) return false; // Check from 2 to n-1 for ( $i = 2; $i < $n ; $i ++) if ( $n % $i == 0) return false; return true; } // Driver Code $tet = isPrime(11) ? " true" : " false" ; echo $tet ; $tet = isPrime(15) ? " true" : " false" ; echo $tet ; // This code is contributed by m_kit ?> |
Javascript
<script> // A school method based Javascript program to check if a // number is prime function isPrime(n) { // Corner case if (n <= 1) return false ; // Check from 2 to n-1 for (let i = 2; i < n; i++) if (n % i == 0) return false ; return true ; } // Driver Program to test above function isPrime(11)? document.write( " true" + "<br>" ): document.write( " false" + "<br>" ); isPrime(15)? document.write( " true" + "<br>" ): document.write( " false" + "<br>" ); // This code is contributed by Mayank Tyagi </script> |
输出:
truefalse
该解的时间复杂度为O(n) 优化的学校方法 我们可以进行以下优化:
- 我们不必查到n,我们可以查到n√n,因为n的较大因子必须是已经检查过的较小因子的倍数。
- 通过观察所有素数的形式为6k±1(2和3除外),可以进一步改进算法。这是因为对于某些整数k和i=-1、0、1、2、3或4,所有整数都可以表示为(6k+i);2分(6k+0)、(6k+2)、(6k+4);3分(6k+3)。所以一个更有效的方法是测试n是否可以被2或3整除,然后检查表6k±1中的所有数字。(来源: 维基百科 )
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