求第三个数，使三个数之和成为质数
原文:https://www . geeksforgeeks . org/find-third-third-number-so-sum-all-three-number-成为质数/
给定两个数字 A 和 B 。任务是找到大于或等于 1 的最小正整数，使这三个数的和成为素数。 例:
输入: A = 2，B = 3 输出: 2 解释: 第三个数是 2，因为如果我们把这三个数加起来得到的 和是 7，这是一个质数。 输入: A = 5，B = 3 输出: 3
进场:
- 首先在 sum 变量中存储给定 2 个数的和。
- 现在，检查和 1 的位“与”(& ) 是否等于 1(检查数字是否为奇数)。
- 如果等于 1，则将 2 分配给变量温度，并进入步骤 4。
- 否则检查 sum 和 temp 变量的和值是否为素数。如果是质数，则打印 temp 变量的值，否则将 temp 变量加 2，直到它小于质数。
以下是上述方法的实现:
C++
// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
// Function that will check
// whether number is prime or not
bool prime(int n)
{
for (int i = 2; i * i <= n; i++) {
if (n % i == 0)
return false;
}
return true;
}
// Function to print the 3rd number
void thirdNumber(int a, int b)
{
int sum = 0, temp = 0;
sum = a + b;
temp = 1;
// If the sum is odd
if (sum & 1) {
temp = 2;
}
// If sum is not prime
while (!prime(sum + temp)) {
temp += 2;
}
cout << temp;
}
// Driver code
int main()
{
int a = 3, b = 5;
thirdNumber(a, b);
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.util.*;
class GFG
{
// Function that will check
// whether number is prime or not
static boolean prime(int n)
{
for (int i = 2; i * i <= n; i++)
{
if (n % i == 0)
return false;
}
return true;
}
// Function to print the 3rd number
static void thirdNumber(int a, int b)
{
int sum = 0, temp = 0;
sum = a + b;
temp = 1;
// If the sum is odd
if (sum == 0)
{
temp = 2;
}
// If sum is not prime
while (!prime(sum + temp))
{
temp += 2;
}
System.out.print(temp);
}
// Driver code
static public void main (String []arr)
{
int a = 3, b = 5;
thirdNumber(a, b);
}
}
// This code is contributed by 29AjayKumar
Python 3
# Python3 implementation of the above approach
# Function that will check
# whether number is prime or not
def prime(n):
for i in range(2, n + 1):
if i * i > n + 1:
break
if (n % i == 0):
return False
return True
# Function to print the 3rd number
def thirdNumber(a, b):
summ = 0
temp = 0
summ = a + b
temp = 1
#If the summ is odd
if (summ & 1):
temp = 2
#If summ is not prime
while (prime(summ + temp) == False):
temp += 2
print(temp)
# Driver code
a = 3
b = 5
thirdNumber(a, b)
# This code is contributed by Mohit Kumar
C
// C# implementation of the above approach
using System;
class GFG
{
// Function that will check
// whether number is prime or not
static bool prime(int n)
{
for (int i = 2; i * i <= n; i++)
{
if (n % i == 0)
return false;
}
return true;
}
// Function to print the 3rd number
static void thirdNumber(int a, int b)
{
int sum = 0, temp = 0;
sum = a + b;
temp = 1;
// If the sum is odd
if (sum == 0)
{
temp = 2;
}
// If sum is not prime
while (!prime(sum + temp))
{
temp += 2;
}
Console.Write (temp);
}
// Driver code
static public void Main ()
{
int a = 3, b = 5;
thirdNumber(a, b);
}
}
// This code is contributed by Sachin.
java 描述语言
<script>
// Javascript implementation of the above approach
// Function that will check
// whether number is prime or not
function prime(n)
{
for (let i = 2; i * i <= n; i++) {
if (n % i == 0)
return false;
}
return true;
}
// Function to print the 3rd number
function thirdNumber(a, b)
{
let sum = 0, temp = 0;
sum = a + b;
temp = 1;
// If the sum is odd
if ((sum & 1) != 0) {
temp = 2;
}
// If sum is not prime
while (!prime(sum + temp)) {
temp += 2;
}
document.write(temp);
}
let a = 3, b = 5;
thirdNumber(a, b);
// This code is contributed by divyeshrabadiya07.
</script>
Output:
3
时间复杂度:O(sqrt(n))
辅助空间:0(1)
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