Palindromic Squares

最新推荐文章于 2024-06-17 11:41:05 发布

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最新推荐文章于 2024-06-17 11:41:05 发布

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分类专栏： usaco解题报告文章标签： usaco解题报告

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本文介绍了一道编程题目，任务是找出1至300范围内，在特定进制下其平方为回文数的所有整数，并提供了Python实现代码。

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题目描述：

Palindromes are numbers that read the same forwards as backwards. The number 12321 is a typical palindrome.

Given a number base B (2 <= B <= 20 base 10), print all the integers N (1 <= N <= 300 base 10) such that the square of N is palindromic when expressed in base B; also print the value of that palindromic square. Use the letters 'A', 'B', and so on to represent the digits 10, 11, and so on.

Print both the number and its square in base B.

输入输出格式：

INPUT FORMAT

A single line with B, the base (specified in base 10).

SAMPLE INPUT (file palsquare.in)

OUTPUT FORMAT

Lines with two integers represented in base B. The first integer is the number whose square is palindromic; the second integer is the square itself. NOTE WELL THAT BOTH INTEGERS ARE IN BASE B!

SAMPLE OUTPUT (file palsquare.out)

题目大意：

输入一个（2-20）进制数，判断这个数（1-300范围）的平方是不是回文数，所回文数是指121，从左到右和从右到左是一样的数，输入的哪一行是数的进制，

解题思路：

这是一道水题，注意一点就行，10进制转任意进制用辗转相除法，具体过程看我代码的f（x，n）外带我用python写的，基本没啥解题思路，直接暴力就完事，解题代码如下

"""
ID: scwswx2
LANG: PYTHON3
TASK: palsquare
"""
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','A','B','C','D','E','F','G','H','I','J']
    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('palsquare.in','r')
fout=open('palsquare.out','w')
a=int(fin.readline().strip('\n'))#输入的进制数
for i in range(1,300):
    before=f(i,a)
    palsquareBefor=f(int(before,a)*int(before,a),a)
    palsquareAfter=palsquareBefor[::-1]
    if palsquareBefor == palsquareAfter:
        #print(before,palsquareBefor)
        fout.write(before+' '+palsquareBefor+'\n')
fout.close()