计算具有所有素数的给定矩阵中的行数和列数

给定一个大小为N*M的二维矩阵arr[] ，任务是找出所有素数的行数和列数。

例子：

Input: arr[]= { { 2, 5, 7 }, { 3, 10, 4 }, { 11, 13, 17 } };
Output: 3
Explanation: 
2 Rows: {2, 5, 7}, {11, 13, 17}
1 Column: {2, 3, 11}

Input: arr[]={ { 1, 4 }, { 4, 6 } }
Output: 0

方法：按照以下步骤解决此问题：

1. 应用 Sieve 算法找出所有素数。
2. 逐行遍历所有元素以找到具有所有素数的行数。
3. 对所有列应用上述步骤。
4. 根据上述观察返回答案。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
#define MAXN 5000
 
bool prime[MAXN + 1];
 
// Sieve to find prime numbers
void SieveOfEratosthenes(int n)
{
    memset(prime, true, sizeof(prime));
 
    for (int p = 2; p * p <= n; p++) {
 
        if (prime[p] == true) {
 
            for (int i = p * p; i <= n; i += p) {
                prime[i] = false;
            }
        }
    }
}
 
// Function to find the number of rows
// and columns having all primes
int primeRowCol(vector >& arr)
{
    int N = arr.size();
    int M = arr[0].size();
 
    SieveOfEratosthenes(MAXN);
 
    int cnt = 0;
 
    // Counting Rows with all primes
    for (int i = 0; i < N; i++) {
 
        bool flag = 1;
        for (int j = 0; j < M; ++j) {
            if (!prime[arr[i][j]]) {
                flag = 0;
                break;
            }
        }
        if (flag) {
            cnt += 1;
        }
    }
 
    // Counting Cols with all primes
    for (int i = 0; i < M; i++) {
 
        bool flag = 1;
        for (int j = 0; j < N; ++j) {
            if (!prime[arr[j][i]]) {
                flag = 0;
                break;
            }
        }
        if (flag) {
            cnt += 1;
        }
    }
 
    return cnt;
}
 
// Driver Code
int main()
{
 
    vector > arr
        = { { 2, 5, 7 }, { 3, 10, 4 }, { 11, 13, 17 } };
    cout << primeRowCol(arr);
}

Java

// Java program for the above approach
class GFG {
 
  static int MAXN = 5000;
  static boolean[] prime = new boolean[MAXN + 1];
 
  // Sieve to find prime numbers
  static void SieveOfEratosthenes(int n) {
    for (int i = 0; i < MAXN; i++) {
      prime[i] = true;
    }
 
    for (int p = 2; p * p <= n; p++) {
 
      if (prime[p] == true) {
 
        for (int i = p * p; i <= n; i += p) {
          prime[i] = false;
        }
      }
    }
  }
 
  // Function to find the number of rows
  // and columns having all primes
  static int primeRowCol(int[][] arr) {
    int N = arr.length;
    int M = arr[0].length;
 
    SieveOfEratosthenes(MAXN);
 
    int cnt = 0;
 
    // Counting Rows with all primes
    for (int i = 0; i < N; i++) {
 
      boolean flag = true;
      for (int j = 0; j < M; ++j) {
        if (!prime[arr[i][j]]) {
          flag = false;
          break;
        }
      }
      if (flag) {
        cnt += 1;
      }
    }
 
    // Counting Cols with all primes
    for (int i = 0; i < M; i++) {
 
      boolean flag = true;
      for (int j = 0; j < N; ++j) {
        if (!prime[arr[j][i]]) {
          flag = false;
          break;
        }
      }
      if (flag) {
        cnt += 1;
      }
    }
 
    return cnt;
  }
 
  // Driver Code
  public static void main(String args[]) {
 
    int[][] arr = { { 2, 5, 7 }, { 3, 10, 4 }, { 11, 13, 17 } };
    System.out.println(primeRowCol(arr));
  }
}
 
// This code is contributed by gfgking

Python3

# Python 3 program for the above approach
MAXN = 5000
 
prime = [True]*(MAXN + 1)
 
# Sieve to find prime numbers
def SieveOfEratosthenes(n):
 
    p = 2
    while p * p <= n:
 
        if (prime[p] == True):
 
            for i in range(p * p, n + 1, p):
                prime[i] = False
 
        p += 1
 
# Function to find the number of rows
# and columns having all primes
def primeRowCol(arr):
 
    N = len(arr)
    M = len(arr[0])
 
    SieveOfEratosthenes(MAXN)
 
    cnt = 0
 
    # Counting Rows with all primes
    for i in range(N):
 
        flag = 1
        for j in range(M):
            if (not prime[arr[i][j]]):
                flag = 0
                break
 
        if (flag):
            cnt += 1
 
    # Counting Cols with all primes
    for i in range(M):
 
        flag = 1
        for j in range(N):
            if (not prime[arr[j][i]]):
                flag = 0
                break
 
        if (flag):
            cnt += 1
 
    return cnt
 
# Driver Code
if __name__ == "__main__":
 
    arr = [[2, 5, 7], [3, 10, 4], [11, 13, 17]]
    print(primeRowCol(arr))
 
    # This code is contributed by ukasp.

C#

// C# program for the above approach
using System;
class GFG
{
 
static int MAXN = 5000;
static bool []prime = new bool[MAXN + 1];
 
// Sieve to find prime numbers
static void SieveOfEratosthenes(int n)
{
    for(int i = 0; i < MAXN; i++){
        prime[i] = true;
    }
 
    for (int p = 2; p * p <= n; p++) {
 
        if (prime[p] == true) {
 
            for (int i = p * p; i <= n; i += p) {
                prime[i] = false;
            }
        }
    }
}
 
// Function to find the number of rows
// and columns having all primes
static int primeRowCol(int [,]arr)
{
    int N = arr.GetLength(0);
    int M = arr.GetLength(1);
 
    SieveOfEratosthenes(MAXN);
 
    int cnt = 0;
 
    // Counting Rows with all primes
    for (int i = 0; i < N; i++) {
 
        bool flag = true;
        for (int j = 0; j < M; ++j) {
            if (!prime[arr[i, j]]) {
                flag = false;
                break;
            }
        }
        if (flag) {
            cnt += 1;
        }
    }
 
    // Counting Cols with all primes
    for (int i = 0; i < M; i++) {
 
        bool flag = true;
        for (int j = 0; j < N; ++j) {
            if (!prime[arr[j, i]]) {
                flag = false;
                break;
            }
        }
        if (flag) {
            cnt += 1;
        }
    }
 
    return cnt;
}
 
// Driver Code
public static void Main()
{
 
    int [,]arr
        = { { 2, 5, 7 }, { 3, 10, 4 }, { 11, 13, 17 } };
    Console.Write(primeRowCol(arr));
}
}
 
// This code is contributed by Samim Hosdsain Mondal.

Javascript

      // JavaScript code for the above approach
      let MAXN = 5000
      let prime = new Array(MAXN + 1).fill(true);
 
      // Sieve to find prime numbers
      function SieveOfEratosthenes(n)
      {
          for (let p = 2; p * p <= n; p++)
          {
              if (prime[p] == true) {
 
                  for (let i = p * p; i <= n; i += p) {
                      prime[i] = false;
                  }
              }
          }
      }
 
      // Function to find the number of rows
      // and columns having all primes
      function primeRowCol(arr) {
          let N = arr.length;
          let M = arr[0].length;
 
          SieveOfEratosthenes(MAXN);
 
          let cnt = 0;
 
          // Counting Rows with all primes
          for (let i = 0; i < N; i++) {
 
              let flag = 1;
              for (let j = 0; j < M; ++j) {
                  if (!prime[arr[i][j]]) {
                      flag = 0;
                      break;
                  }
              }
              if (flag) {
                  cnt += 1;
              }
          }
 
          // Counting Cols with all primes
          for (let i = 0; i < M; i++) {
 
              let flag = 1;
              for (let j = 0; j < N; ++j) {
                  if (!prime[arr[j][i]]) {
                      flag = 0;
                      break;
                  }
              }
              if (flag) {
                  cnt += 1;
              }
          }
 
          return cnt;
      }
 
      // Driver Code
      let arr = [[2, 5, 7], [3, 10, 4], [11, 13, 17]];
      document.write(primeRowCol(arr));
 
// This code is contributed by Potta Lokesh
  

输出

3

时间复杂度： O(N*M)
辅助空间： O(max(arr))