给定一个包含n个不同数字的数组。任务是找到阵列中不存在的最小素数。
注意：如果在数组的最大元素前没有素数，请打印“不缺少素数”。
例子：
Input: arr[] = {9, 11, 4, 2, 3, 7, 0, 1}
Output: 5
5 is the smallest prime, which is not present in array.
Input: arr[] = {3, 0, 2, 5}
Output: No prime number missing
As 5 is the maximum element and all prime numbers upto 5
are present in the array.
方法：首先，使用Eratosthenes筛子查找所有素数，然后顺序检查那里不存在哪个素数。只需遍历数组并使用哈希检查数字是否存在。
下面是上述方法的实现：
C++
// C++ implementation of the above approach
#include
using namespace std;
#define ll long long int
// this store all prime number
// upto 10^5
// this function find all prime
vector findPrime(int MAX)
{
bool pm[MAX + 1];
memset(pm, true, sizeof(pm));
// use sieve to find prime
pm[0] = pm[1] = false;
for (int i = 2; i <= MAX; i++)
if (pm[i])
for (int j = 2 * i; j <= MAX; j += i)
pm[j] = false;
// if number is prime then
// store it in prime vector
vector prime;
for (int i = 0; i <= MAX; i++)
if (pm[i])
prime.push_back(i);
return prime;
}
// Function to find the smallest prime missing
int findSmallest(int arr[], int n)
{
int MAX = *max_element(arr, arr + n);
// first of all find all prime
vector prime = findPrime(MAX);
// now store all number as index of freq arr
// so that we can improve searching
unordered_set s;
for (int i = 0; i < n; i++)
s.insert(arr[i]);
// now check for each prime
int ans = -1;
for (int i = 0; i < prime.size(); i++)
if (s.find(prime[i]) == s.end()) {
ans = prime[i];
break;
}
return ans;
}
// Driver code
int main()
{
int arr[] = { 3, 0, 1, 2, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
// find smallest prime
// which is not present
if (findSmallest(arr, n) == -1)
cout << "No prime number missing";
else
cout << findSmallest(arr, n);
return 0;
} Java
// Java implementation of the above approach
import java.util.*;
class GFG {
// this store all prime number
// upto 10^5
// this function find all prime
static Vector findPrime(int MAX)
{
boolean pm[] = new boolean[MAX + 1];
for (int i = 0; i < pm.length; i++)
pm[i] = true;
// use sieve to find prime
pm[0] = pm[1] = false;
for (int i = 2; i <= MAX; i++)
if (pm[i])
for (int j = 2 * i; j <= MAX; j += i)
pm[j] = false;
// if number is prime then
// store it in prime vector
Vector prime = new Vector();
for (int i = 0; i <= MAX; i++)
if (pm[i])
prime.add(i);
return prime;
}
static int max_element(int arr[])
{
int max = arr[0];
for (int i = 0; i < arr.length; i++)
max = Math.max(max, arr[i]);
return max;
}
// Function to find the smallest prime missing
static int findSmallest(int arr[], int n)
{
int MAX = max_element(arr);
// first of all find all prime
Vector prime = findPrime(MAX);
// now store all number as index of freq arr
// so that we can improve searching
Set s = new HashSet();
for (int i = 0; i < arr.length; i++)
s.add(arr[i]);
// now check for each prime
long ans = -1;
for (int i = 0; i < prime.size(); i++) {
if (!s.contains(prime.get(i))) {
ans = (prime.get(i));
break;
}
}
return (int)ans;
}
// Driver code
public static void main(String args[])
{
int arr[] = { 3, 0, 1, 2, 7 };
int n = arr.length;
// find smallest prime
// which is not present
if (findSmallest(arr, n) == -1)
System.out.print("No prime number missing");
else
System.out.print(findSmallest(arr, n));
}
}
// This code is contributed by Arnab Kundu Python3
# Python3 implementation of the above approach
# This function finds all
# prime numbers upto 10 ^ 5
def findPrime(MAX):
pm = [True] * (MAX + 1)
# use sieve to find prime
pm[0], pm[1] = False, False
for i in range(2, MAX + 1):
if pm[i] == True:
for j in range(2 * i, MAX + 1, i):
pm[j] = False
# If number is prime, then
# store it in prime list
prime = []
for i in range(0, MAX + 1):
if pm[i] == True:
prime.append(i)
return prime
# Function to find the smallest prime missing
def findSmallest(arr, n):
MAX = max(arr)
# first of all find all prime
prime = findPrime(MAX)
# now store all number as index of freq
# arr so that we can improve searching
s = set()
for i in range(0, n):
s.add(arr[i])
# now check for each prime
ans = -1
for i in range(0, len(prime)):
if prime[i] not in s:
ans = prime[i]
break
return ans
# Driver Code
if __name__ == "__main__":
arr = [3, 0, 1, 2, 7]
n = len(arr)
# find smallest prime
# which is not present
if(findSmallest(arr, n) == -1):
print("No prime number missing")
else:
print(findSmallest(arr, n))
# This code is contributed by Rituraj JainC#
// C# implementation of the above approach
using System;
using System.Collections;
using System.Collections.Generic;
class GFG
{
// this store all prime number
// upto 10^5
// this function find all prime
static ArrayList findPrime(int MAX)
{
bool[] pm = new bool[MAX + 1];
for (int i = 0; i < MAX + 1; i++)
pm[i] = true;
// use sieve to find prime
pm[0] = pm[1] = false;
for (int i = 2; i <= MAX; i++)
if (pm[i])
for (int j = 2 * i; j <= MAX; j += i)
pm[j] = false;
// if number is prime then
// store it in prime vector
ArrayList prime = new ArrayList();
for (int i = 0; i <= MAX; i++)
if (pm[i])
prime.Add(i);
return prime;
}
static int max_element(int []arr)
{
int max = arr[0];
for (int i = 0; i < arr.Length; i++)
max = Math.Max(max, arr[i]);
return max;
}
// Function to find the smallest prime missing
static int findSmallest(int []arr, int n)
{
int MAX = max_element(arr);
// first of all find all prime
ArrayList prime = findPrime(MAX);
// now store all number as index of freq arr
// so that we can improve searching
HashSet s = new HashSet();
for (int i = 0; i < arr.Length; i++)
s.Add(arr[i]);
// now check for each prime
int ans = -1;
for (int i = 0; i < prime.Count; i++)
{
if (!s.Contains((int)prime[i]))
{
ans = (int)(prime[i]);
break;
}
}
return (int)ans;
}
// Driver code
static void Main()
{
int []arr = { 3, 0, 1, 2, 7 };
int n = arr.Length;
// find smallest prime
// which is not present
if (findSmallest(arr, n) == -1)
Console.Write("No prime number missing");
else
Console.Write(findSmallest(arr, n));
}
}
// This code is contributed by mits 输出：
5