PAT 1019 General Palindromic Number (20分) python实现

向前或向后写入时相同的数字称为回文数。例如，1234321是回文数。所有单数数字都是回文数字。虽然回文数在十进制中最常被考虑，但回文性的概念可以适用于任何数字系统中的自然数。考虑基b≥2中的一个数字N>0，其中它是用标准表示法编写的，用k+1位数字(​i​​)作为∑​i=0​k​​(​i​​b​i​​)。在这里，和往常一样，0≤a​i​​ Input Specification:
Each input file contains one test case. Each case consists of two positive numbers N and b, where 0 ​9
​​ is the decimal number and 2≤b≤10
​9
​​ is the base. The numbers are separated by a space.

Output Specification:
For each test case, first print in one line Yes if N is a palindromic number in base b, or No if not. Then in the next line, print N as the number in base b in the form "a
​k
​​ a
​k−1
​​ … a
​0
​​ ". Notice that there must be no extra space at the end of output.

Sample Input 1:
27 2
Sample Output 1:
Yes
1 1 0 1 1
Sample Input 2:
121 5
Sample Output 2:
No

具体思路

就是数制转换，再加上个回文数判断，，很简单，因为数值大，我就用的python

具体代码

def main():
    y=input('').split( )
    ben=int(y[0])
    cha=int(y[1])
    s=[]
    sum=1
    while ben!=0:#基本数制转换方法
        sum=ben%cha
        s.append(sum)
        ben=ben//cha
    j=len(s)-1
    i=0
    biaoji=0
    while j-i>1:#判断是否为回文数
        if(s[j]!=s[i]):
            biaoji=1
        j=j-1
        i=i+1
    if len(s)==2:
        if(s[j]!=s[i]):
            biaoji=1
    if biaoji==1:
        print("No")
    else:
        print("Yes")
    s=s[::-1]#记得逆序输出
    for i in range(len(s)):
        if i!=0:
            print(' ',end='')
        print(str(s[i]),end='')
main()

很简单