检查矩阵和是否为素数

给定一个矩阵mat[][] ，任务是检查矩阵元素的总和是否为素数。
例子：

Input: mat[][] = {{1, 2}, {2, 1}} 
Output: NO 
Explanation: 
Sum of Matrix = 1 + 2 + 2 + 1 = 6 
Since 6 is not prime. Therefore, output is NO
Input: mat[][] = {{1, 2}, {2, 2}} 
Output: YES 
Explanation: 
Sum of matrix = 1 + 2 + 2 + 2 = 7 
Since 7 is prime. Therefore, output is YES 
 

方法：想法是使用两个嵌套循环找到矩阵的和，然后最后检查矩阵的和是否为素数。如果是，则输出为 YES，否则输出为 NO。
下面是上述方法的实现：

C++

// C++ implementation to check
// if the sum of matrix
// is prime or not
 
#include 
using namespace std;
 
const int N = 4, M = 5;
 
// Function to check
// whether a number
// is prime or not
bool isPrime(int n)
{
    // Corner case
    if (n <= 1)
        return false;
 
    // Check from 2 to n-1
    for (int i = 2; i <= sqrt(n); i++)
        if (n % i == 0)
            return false;
 
    return true;
}
 
// Function for to find the sum
// of the given matrix
int takeSum(int a[N][M])
{
    int s = 0;
    for (int i = 0; i < N; i++)
        for (int j = 0; j < M; j++)
            s += a[i][j];
 
    return s;
}
 
// Driver Code
int main()
{
 
    int a[N][M] = { { 1, 2, 3, 4, 2 },
                    { 0, 1, 2, 3, 34 },
                    { 0, 34, 21, 12, 12 },
                    { 1, 2, 3, 6, 6 } };
    int sum = takeSum(a);
 
    if (isPrime(sum))
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
    return 0;
}

Java

// Java implementation to check if
// the sum of matrix is prime or not
class GFG{
 
static int N = 4, M = 5;
 
// Function to check whether
// a number is prime or not
static boolean isPrime(int n)
{
     
    // Corner case
    if (n <= 1)
        return false;
 
    // Check from 2 to n-1
    for(int i = 2; i <= Math.sqrt(n); i++)
       if (n % i == 0)
           return false;
 
    return true;
}
 
// Function for to find the sum
// of the given matrix
static int takeSum(int a[][])
{
    int s = 0;
     
    for(int i = 0; i < N; i++)
       for(int j = 0; j < M; j++)
          s += a[i][j];
 
    return s;
}
 
// Driver Code
public static void main(String[] args)
{
    int a[][] = { { 1, 2, 3, 4, 2 },
                  { 0, 1, 2, 3, 34 },
                  { 0, 34, 21, 12, 12 },
                  { 1, 2, 3, 6, 6 } };
                   
    int sum = takeSum(a);
 
    if (isPrime(sum))
        System.out.print("YES" + "\n");
    else
        System.out.print("NO" + "\n");
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation to check if
# the sum of matrix is prime or not
import math
 
# Function to check whether a number
# is prime or not
def isPrime(n):
 
    # Corner case
    if (n <= 1):
        return False;
 
    # Check from 2 to n-1
    for i in range(2, (int)(math.sqrt(n)) + 1):
        if (n % i == 0):
            return False;
 
    return True;
 
# Function for to find the sum
# of the given matrix
def takeSum(a):
 
    s = 0
    for i in range(0, 4):
        for j in range(0, 5):
            s += a[i][j]
 
    return s;
 
# Driver Code
a = [ [ 1, 2, 3, 4, 2 ],
      [ 0, 1, 2, 3, 34 ],
      [ 0, 34, 21, 12, 12 ],
      [ 1, 2, 3, 6, 6 ] ];
sum = takeSum(a);
 
if (isPrime(sum)):
    print("YES")
else:
    print("NO")
     
# This code is contributed by grand_master

C#

// C# implementation to check if
// the sum of matrix is prime or not
using System;
class GFG{
 
static int N = 4, M = 5;
 
// Function to check whether
// a number is prime or not
static Boolean isPrime(int n)
{
     
    // Corner case
    if (n <= 1)
        return false;
 
    // Check from 2 to n-1
    for(int i = 2; i <= Math.Sqrt(n); i++)
    if (n % i == 0)
        return false;
 
    return true;
}
 
// Function for to find the sum
// of the given matrix
static int takeSum(int [][]a)
{
    int s = 0;
     
    for(int i = 0; i < N; i++)
    for(int j = 0; j < M; j++)
        s += a[i][j];
 
    return s;
}
 
// Driver Code
public static void Main(String[] args)
{
    int [][]a = new int[][]
                {
                    new int[] { 1, 2, 3, 4, 2 },
                    new int[] { 0, 1, 2, 3, 34 },
                    new int[] { 0, 34, 21, 12, 12 },
                    new int[] { 1, 2, 3, 6, 6 }
                };
                 
    int sum = takeSum(a);
 
    if (isPrime(sum))
        Console.Write("YES" + "\n");
    else
        Console.Write("NO" + "\n");
}
}
 
// This code is contributed by shivanisinghss2110

Javascript

    // Javascript implementation to check
    // if the sum of matrix
    // is prime or not
     
    let N = 4, M = 5;
   
    // Function to check
    // whether a number
    // is prime or not
    function isPrime(n)
    {
        // Corner case
        if (n <= 1)
            return false;
 
        // Check from 2 to n-1
        for (let i = 2; i <= Math.sqrt(n); i++)
            if (n % i == 0)
                return false;
 
        return true;
    }
 
    // Function for to find the sum
    // of the given matrix
    function takeSum(a)
    {
        let s = 0;
        for (let i = 0; i < N; i++)
            for (let j = 0; j < M; j++)
                s += a[i][j];
 
        return s;
    }
 
    let a = [ [ 1, 2, 3, 4, 2 ],
             [ 0, 1, 2, 3, 34 ],
             [ 0, 34, 21, 12, 12 ],
             [ 1, 2, 3, 6, 6 ] ];
    let sum = takeSum(a);
   
    if (isPrime(sum))
        document.write("YES");
    else
        document.write("NO");
     

输出：

YES

时间复杂度： O(N*M)

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