一个数奇数位数字之和的素性检验
给定一个整数“n”，任务是检查奇数位置(从右到左)的数字总和是否为素数。 如果是质数，则打印“是”或“否”。 例:
输入:n = 123 T3】输出: NO As，1 + 3 = 4 不是素数。 输入: n = 42 输出: YES 既然，2 是素数。
逼近:首先求奇数位即 1，3，5，…(从右开始)的位数之和。 如果总和是质数，则打印“是”，否则打印“否”。 以下是上述办法的实施:
C++
// C++ program to do Primality test
// for the sum of digits at
// odd places of a number
#include <bits/stdc++.h>
using namespace std;
// Function that return sum
// of the digits at odd places
int sum_odd(int n)
{
int sum = 0, pos = 1;
while (n) {
if (pos % 2 == 1)
sum += n % 10;
n = n / 10;
pos++;
}
return sum;
}
// Function that returns true
// if the number is prime
// else false
bool check_prime(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This condition is checked so that
// we can skip middle five
// numbers in the below loop
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
// Driver code
int main()
{
int n = 223;
// Get the sum of the
// digits at odd places
int sum = sum_odd(n);
if (check_prime(sum))
cout << "YES" << endl;
else
cout << "NO" << endl;
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to do Primality test
// for the sum of digits at
// odd places of a number
import java.io.*;
class GFG {
// Function that return sum
// of the digits at odd places
static int sum_odd(int n)
{
int sum = 0, pos = 1;
while (n>0) {
if (pos % 2 == 1)
sum += n % 10;
n = n / 10;
pos++;
}
return sum;
}
// Function that returns true
// if the number is prime
// else false
static boolean check_prime(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This condition is checked so that
// we can skip middle five
// numbers in the below loop
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
// Driver code
public static void main (String[] args) {
int n = 223;
// Get the sum of the
// digits at odd places
int sum = sum_odd(n);
if (check_prime(sum))
System.out.println ("YES" );
else
System.out.println("NO");
}
}
Python 3
# Python3 program to do Primality test
# for the sum of digits at
# odd places of a number
# Function that return sum
# of the digits at odd places
def sum_odd(n):
sums = 0
pos = 1
while (n!=0):
if (pos % 2 == 1):
sums += n % 10
n = n // 10
pos+=1
return sums
# Function to check if a
# number is prime
def check_prime(n):
# Corner cases
if (n <= 1):
return False
if (n <= 3):
return True
# This is checked so that we can skip
# middle five numbers in below loop
if (n % 2 == 0 or n % 3 == 0):
return False
for i in range(5,n,6):
if (n % i == 0 or n % (i + 2) == 0):
return False
return True
#driver code
n = 223
# Get the sum of the
# digits at odd places
sums = sum_odd(n)
if (check_prime(sums)):
print("YES")
else:
print("NO")
#this code is improved by sahilshelangia
C
// C# program to do Primality test
// for the sum of digits at
// odd places of a number
using System;
public class GFG{
// Function that return sum
// of the digits at odd places
static int sum_odd(int n)
{
int sum = 0, pos = 1;
while (n>0) {
if (pos % 2 == 1)
sum += n % 10;
n = n / 10;
pos++;
}
return sum;
}
// Function that returns true
// if the number is prime
// else false
static bool check_prime(int n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This condition is checked so that
// we can skip middle five
// numbers in the below loop
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
// Driver code
static public void Main (){
int n = 223;
// Get the sum of the
// digits at odd places
int sum = sum_odd(n);
if (check_prime(sum))
Console.WriteLine("YES" );
else
Console.WriteLine("NO");
}
}
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to do Primality test
// for the sum of digits at odd
// places of a number
// Function that return sum
// of the digits at odd places
function sum_odd($n)
{
$sum = 0;
$pos = 1;
while ($n)
{
if ($pos % 2 == 1)
$sum += $n % 10;
$n = (int)($n / 10);
$pos++;
}
return $sum;
}
// Function that returns true
// if the number is prime
// else false
function check_prime($n)
{
// Corner cases
if ($n <= 1)
return false;
if ($n <= 3)
return true;
// This condition is checked so
// that we can skip middle five
// numbers in the below loop
if ($n % 2 == 0 || $n % 3 == 0)
return false;
for ($i = 5; $i * $i <= $n;
$i = ($i + 6))
if ($n % $i == 0 ||
$n % ($i + 2) == 0)
return false;
return true;
}
// Driver code
$n = 223;
// Get the sum of the
// digits at odd places
$sum = sum_odd($n);
if (check_prime($sum))
echo "YES";
else
echo "NO";
// This code is contributed by ajit
?>
java 描述语言
<script>
// JavaScript program to do Primality test
// for the sum of digits at
// odd places of a number
// Function that return sum
// of the digits at odd places
function sum_odd(n)
{
let sum = 0, pos = 1;
while (n) {
if (pos % 2 == 1)
sum += n % 10;
n = Math.floor(n / 10);
pos++;
}
return sum;
}
// Function that returns true
// if the number is prime
// else false
function check_prime(n)
{
// Corner cases
if (n <= 1)
return false;
if (n <= 3)
return true;
// This condition is checked so that
// we can skip middle five
// numbers in the below loop
if (n % 2 == 0 || n % 3 == 0)
return false;
for (let i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
// Driver code
let n = 223;
// Get the sum of the
// digits at odd places
let sum = sum_odd(n);
if (check_prime(sum))
document.write("YES" + "<br>");
else
document.write("NO" + "<br>");
// This code is contributed by Surbhi Tyagi.
</script>
Output:
YES
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