题解 | #素数回文#

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<math.h>
int is_single(long long n)
{
	int i = 0;
	int j = 0;
	for (i = 2; i < n; i++)
	{
		if (n % i == 0)
		{
			return 0;
		}
	}
	//只由1-9组成，并且每个数只出现一次，如13,23,1289。
	int arr[9] = { 0 };
	int count = 0;
	int m = n;
	i = 0;
	while (m)
	{
		arr[i] = m % 10;
		m /= 10;
		i++;
		count++;
	}
	for (i = 0; i < count; i++)
	{
		for (j = 0; j < i; j++)
		{
			if (arr[i] == arr[j])
			{
				return 0;
			}
		}
	}
	int crease = 0;
	for (i = 0; i < count - 1; i++)
	{
		//位数从高到低为递减或递增，如2459，87631。
		//计算
		if (arr[i] < arr[i + 1])
		{
			crease++;
		}
		else if (arr[i] > arr[i + 1])
		{
			crease++;
		}
	}
	if (crease == 0 || crease == count - 1)
	{
		return 1;
	}
	else
	{
		return 0;
	}
}
long long check1(long long n)
{
	if (is_single(n))
	{
		//开始回文
		int num = 0;
		long long sum1 = n;
		while (sum1)
		{
			sum1 /= 10;
			num++;
		}
		long long sum2 = n;
		for (; num > 1; num--)
		{
			sum2 = (sum2 * 10 + n / 10 % 10);
			n /= 10;
		}
		return sum2;
	}
	else
	{
		return 0;
	}
}
int check2(long long n)
//定义一个函数用来判断素数
{
	long long a;
	int b = 0;
	if (n % 2 == 0)
	{
		return 0;
	}
	long double c = sqrt(n);
	for (a = 3; a < c; a += 2)
	{
		if ((n % a) == 0)
		{
			b++;
			break;
		}
	}
	if (b == 0)
	{
		return 1;
	}
	else
	{
		return 0;
	}
}
int main()
{
	long long n = 0;
	scanf("%lld", &n);
	n = check1(n);
	if (n == 0)
	{
		printf("搞错啦！搞错啦！！！\n");
		return 1;
	}
	n = check2(n);
	if (n)
	{
		printf("prime\n");
	}
	else
	{
		printf("noprime\n");
	}
	return 0;
}