小南解题-特殊回文数--57ms，48ms，53ms

#最简单，不超时 -----  57ms

n = int(input())

my_list = []

for i in range(100,1000) :

# 如果该数字是6位数

    if sum(map(int,str(i) + str(i)[::-1])) == n : # 同时判断是否是回文数和累加等于n

        my_list.append(str(i) + str(i)[::-1])

    # 如果该数字是5位数

    if sum(map(int,str(i) + str(i)[:2][::-1])) == n :

        my_list.append(str(i) + str(i)[:2][::-1])

for i in sorted(map(int,my_list)) : # 排序

    print(i)

'''

#法一：参考代码---48ms

n = eval(input())

if n % 2 == 1:

    for a in range(1, 10):

        for b in range(10):

            for c in range(10):

                if (a + b) * 2 + c == n:

                    res = str(a) + str(b) + str(c)

                    print(res + res[1::-1])

else:

    for a in range(1, 10):

        for b in range(10):

            for c in range(10):

                if (a + b) * 2 + c == n:

                    res = str(a) + str(b) + str(c)

                    print(res + res[1::-1])

    for a in range(1, 10):

        for b in range(10):

            for c in range(10):

                if (a + b + c) * 2 == n:

                    res = str(a) + str(b) + str(c)

                    print(res + res[::-1])

#法二：自己，新思路----53ms

n=int(input())

a=[]

for i in range(10,100):

    for j  in range(10):        

        i=str(i)

        if (int(i[0])+int(i[1]))*2+j==n:

            print(i+str(j)+i[::-1])

for i in range(100,1000):      

    i=str(i)

    if (int(i[0])+int(i[1])+int(i[2]))*2==n:

        print(i+i[::-1])

'''