给定数字N，任务是找到小于或等于给定数字N的最大数字，以便在重新排列其数字时可以成为质数。
例子：
Input : N = 99
Output : 98
Explanation : We can rearrange the digits of
98 to 89 and 89 is a prime number.
Input : N = 84896
Output : 84896
Explanation : We can rearrange the digits of
84896 to 46889 which is a prime number.
下面是找到最大数量num <= N的算法，以便可以重新排列num的位数以获得质数：
预处理步骤：生成所有小于或等于给定数字N的质数的列表。可以使用Eratosthenes筛子有效地完成此过程。
主要步骤：主要思想是检查从N到1的所有数字，如果可以重新组合任何数字以形成质数。找到的第一个这样的数字就是答案。
为此，请对每个数字从N到1循环运行：
- 提取给定数字的数字并将其存储在向量中。
- 对该向量进行排序，以获得可以使用这些数字形成的最小数字。
- 对于此向量的每个排列，我们将形成一个数字并检查形成的数字是否为质数。在这里，我们利用预处理步骤。
- 如果是素数，则我们停止循环，这就是我们的答案。
下面是上述方法的实现：
C++
// C++ Program to find a number less than
// or equal to N such that rearranging
// the digits gets a prime number
#include
using namespace std;
// Preprocessed vector to store primes
vector prime(1000001, true);
// Function to generate primes using
// sieve of eratosthenese
void sieve()
{
// Applying sieve of Eratosthenes
prime[0] = prime[1] = false;
for (long long i = 2; i * i <= 1000000; i++) {
if (prime[i]) {
for (long long j = i * i; j <= 1000000; j += i)
prime[j] = false;
}
}
}
// Function to find a number less than
// or equal to N such that rearranging
// the digits gets a prime number
int findNumber(int n)
{
vector v;
bool flag = false;
int num;
// Run loop from n to 1
for (num = n; num >= 1; num--) {
int x = num;
// Clearing the vector
v.clear();
// Extracting the digits
while (x != 0) {
v.push_back(x % 10);
x /= 10;
}
// Sorting the vector to make smallest
// number using digits
sort(v.begin(), v.end());
// Check all permutation of current number
// for primality
while (1) {
long long w = 0;
// Traverse vector to for number
for (auto u : v)
w = w * 10 + u;
// If prime exists
if (prime[w]) {
flag = true;
break;
}
if (flag)
break;
// generating next permutation of vector
if (!next_permutation(v.begin(), v.end()))
break;
}
if (flag)
break;
}
// Required number
return num;
}
// Driver Code
int main()
{
sieve();
int n = 99;
cout << findNumber(n) << endl;
n = 84896;
cout << findNumber(n) << endl;
return 0;
} Python3
# Python 3 Program to find a number less than
# or equal to N such that rearranging
# the digits gets a prime number
from math import sqrt
def next_permutation(a):
# Generate the lexicographically
# next permutation inplace.
# https://en.wikipedia.org/wiki/Permutation
# Generation_in_lexicographic_order
# Return false if there is no next permutation.
# Find the largest index i such that
# a[i] < a[i + 1]. If no such index exists,
# the permutation is the last permutation
for i in reversed(range(len(a) - 1)):
if a[i] < a[i + 1]:
break # found
else: # no break: not found
return False # no next permutation
# Find the largest index j greater than i
# such that a[i] < a[j]
j = next(j for j in reversed(range(i + 1, len(a)))
if a[i] < a[j])
# Swap the value of a[i] with that of a[j]
a[i], a[j] = a[j], a[i]
# Reverse sequence from a[i + 1] up to and
# including the final element a[n]
a[i + 1:] = reversed(a[i + 1:])
return True
# Preprocessed vector to store primes
prime = [True for i in range(1000001)]
# Function to generate primes using
# sieve of eratosthenese
def sieve():
# Applying sieve of Eratosthenes
prime[0] = False
prime[1] = False
for i in range(2,int(sqrt(1000000)) + 1, 1):
if (prime[i]):
for j in range(i * i, 1000001, i):
prime[j] = False
# Function to find a number less than
# or equal to N such that rearranging
# the digits gets a prime number
def findNumber(n):
v = []
flag = False
# Run loop from n to 1
num = n
while(num >= 1):
x = num
v.clear()
# Extracting the digits
while (x != 0):
v.append(x % 10)
x = int(x / 10)
# Sorting the vector to make smallest
# number using digits
v.sort(reverse = False)
# Check all permutation of current number
# for primality
while (1):
w = 0
# Traverse vector to for number
for u in v:
w = w * 10 + u
# If prime exists
if (prime[w]):
flag = True
break
if (flag):
break
# generating next permutation of vector
if (next_permutation(v) == False):
break
if (flag):
break
num -= 1
# Required number
return num
# Driver Code
if __name__ == '__main__':
sieve()
n = 99
print(findNumber(n))
n = 84896
print(findNumber(n))
# This code is contributed by
# Surendra_Gangwar输出：
98
84896