回文数的四种求法

1.基础求法

#include 
int main()
{
    int n;
    printf("Please input a five-digit number\n");
    scanf("%d", &n);
    int a = n % 10;
    int b = n / 10 % 10;
    int c = n / 100 % 10;
    int d = n / 1000 % 10;
    int e = n / 10000 % 10;
    if (a == e && b == d)
        printf("YES!");
    else
        printf("NO!");
}

2.进阶之数组法

#include 
int main()
{
    int a[10000];
    int n;
    int i;
    printf("您将输入几位数？");
    scanf("%d", &n);
    printf("请输入");
    for (i = 0; i < n; i++)
    {
        scanf("%d", &a[i]);
    }
    for (i = 0; i < n; i++)
    {
        if (a[i] != a[n - i - 1])
        {
            printf("NO");
            return 0;
        }
    }
    printf("YES");
}

3.将数字倒序重排法

#include 
int main()
{
    int n;
    int sum = 0;
    scanf("%d", &n);
    for (int i = n; i > 0; i /= 10)
    {
        sum = sum * 10 + i % 10; //把数字倒序
    }
    if (sum == n)
        printf("YES！");
    else
        printf("NO");
}

4.字符串

#include 
#include 
#define N 80
int fun(char *str)
{
    int i, n;
    n = strlen(str); //获得字符串的长度
    for (i = 0; i < n; i++)
    {
        if (str[i] != str[n - 1 - i])
        {
            return 0;
        }
    }
    return 1;
}
main()
{
    char s[N];
    printf("Enter a string: ");
    gets(s);
    if (fun(s))
        printf("  YES\n");
    else
        printf("  NO\n");
}