Dual Palindromes

最新推荐文章于 2020-04-10 16:47:19 发布

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本篇介绍了一个编程问题的解决方法，即找出大于指定数值的前N个数，在2到10进制中至少有两种进制下是回文数。文章提供了Python实现代码，展示了如何将十进制数转换为不同进制，并检查其是否为回文。

题目描述：

A number that reads the same from right to left as when read from left to right is called a palindrome. The number 12321 is a palindrome; the number 77778 is not. Of course, palindromes have neither leading nor trailing zeroes, so 0220 is not a palindrome.

The number 21 (base 10) is not palindrome in base 10, but the number 21 (base 10) is, in fact, a palindrome in base 2 (10101).

Write a program that reads two numbers (expressed in base 10):

N (1 <= N <= 15)
S (0 < S < 10000)

and then finds and prints (in base 10) the first N numbers strictly greater than S that are palindromic when written in two or more number bases (2 <= base <= 10).

Solutions to this problem do not require manipulating integers larger than the standard 32 bits.

输入输出格式：

INPUT FORMAT

A single line with space separated integers N and S.

SAMPLE INPUT (file dualpal.in)

3 25

OUTPUT FORMAT

N lines, each with a base 10 number that is palindromic when expressed in at least two of the bases 2..10. The numbers should be listed in order from smallest to largest.

SAMPLE OUTPUT (file dualpal.out)

26
27
28

题目大意：

输入一个N，S，输出大于S的前n个（1-10）进制至少有2种进制相同的那个数，21的二进制数就是一个回文数

解题思路：

直接按条件搜索就行，不会出现超时的问题，亲测，问题在于一定要注意循环条件，我被坑了4-5次，就是

for i in range(2,12):这里，一定要注意

"""
ID: scwswx2
LANG: PYTHON3
TASK: dualpal
"""
def f(n,x):
    #n为待转换的十进制数，x为进制，取值为2-20
    a=['0','1','2','3','4','5','6','7','8','9','0']
    b=[]
    wx=''
    while True:
        s=n//x#商
        y=n%x#余数
        b=b+[y]
        if s==0:
            break
        n=s
    b.reverse()
    for i in b:
        #print(a[i],end='')
        wx=wx+str(a[i])
    #print(wx)
    return wx
fin=open('dualpal.in','r')
fout=open('dualpal.out','w')
inputA,inputB=map(int,fin.readline().split())#输入的N,S
count = 0#记录找到了几个数字
while count<inputA:
    isdual=0
    inputB = inputB + 1
    for i in range(2,12):
        if isdual>=2:
            #print(inputB)
            fout.write(str(inputB)+'\n')
            count=count+1
            break
        if i>=10:
            before=str(inputB)
        before=f(inputB,i)
        after=before[::-1]
        if before==after:
            isdual=isdual+1
fout.close()