Determine whether a number is a symmetric number (array/non-array solution)

Non-array method:

Core: Turn a number from left to right upside down (reverse order) into a new number, and judge whether it is the same as the original number

Difficulty: Flashback output cannot meet the requirement of forming new numbers

scanf("%d",&n);
while(n)
{
print("%d",n%10);
n/10
}

 Solution: After finding each digit, multiply by 10 and keep shifting to the left (operation)

For example, 1234. (See above for the operation process)

Find 4, after the operation , become 40, add the next digit (3), become 43

After the operation , it becomes 430, plus the next digit (2), becomes 432

After the operation , it becomes 4320, plus the next digit (1), becomes 4321. (Complete the new number)

Specific code:

#include<stdio.h>
int main()
{
	int mx = 0;//mx为旧数  1
	scanf("%d", &mx);
	int m = mx;//备份一份旧数，最后与新数做比较  2 
	int n = 0;;//新数的创建  3  
	while (m)
	{
		int x = m % 10;//为了不改变m的值，创建一个x来接收每一位   4 
		n = n * 10 + x;
		m/=10;
	}
	if (n == mx)
	{
		printf("%d是对称数 ", mx);
	}
	else
			printf("想得美");
				return 0;
}

Note: In order to keep the original value unchanged in the code, a new variable is created for backup (2, 4)

Array method:

Core: store a number in an array, and judge by comparing the first and last elements one by one

Difficulties: 1. To control the cycle of taking the position of the array, and at the same time control the cycle of taking out all the digits of the number

           Solution: Add an if in the for loop to achieve dual control

           2. In the array, the head element head and the tail element hail are judged one by one (hail--), and the selection of the termination condition

           Solution: take a special position, 0 1 2. 2/2=1. 0 1 2 3. 3/2=1.

#include<stdio.h>
int main()
{
	int a,tail ,head , i, arr[1000];
	int flag = 1;//设置判断符，不更改为0，即是对称数  1
	scanf("%d", &a);
	int ma = a;//备份
	if (a< 10)//对称数起码是二位数
	{
		return 0;
	}
	for (i = 0; i < 1000; i++)//可以实现，数组用多少（位）拿多少
	{
		arr[i] = a % 10;
		a /= 10;
		if (a == 0) //循环跳出条件，此时for循环中有两个控制条件
			break;
	}
	for (tail = i,head = 0; head <= i / 2; head++)
	{
		if (arr[head] != arr[tail])
		{
			flag = 0;//若更改为0，不是对称数  2
			break;
		}
		tail--;
	}
	if (flag == 1)
		printf("%d是对称数", ma);
	else
		printf("你在想什么？");
	return 0;
}

Non-array method: method two

(to be continued)