USACO----Palindromic Squares

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于 2018-03-19 16:46:41 发布

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本文介绍了一个程序，该程序在给定的数制下找出所有其平方为回文数的整数（1到300之间）。回文数是指正读和反读都一样的数。文章提供了完整的C++代码实现，并解释了如何将十进制数转换为指定进制的字符串形式，以及如何判断一个数是否为回文数。

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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.

PROGRAM NAME: palsquare

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)

/*
ID: 76875131
PROG: palsquare
LANG: C++11
*/
#include <iostream>
#include <fstream>
#include <string>
#include <algorithm>

using namespace std;

void GetBaseValue(int iValue, int iBase, string &str)
{
   while(iValue)
   {
      int i = iValue % iBase;
      if(i >= 10) str.push_back('A' + i - 10);
      else str.push_back('0' + i);

      iValue /= iBase;
   }
   reverse(str.begin(), str.end());
}

bool IsPalindromic(const string &str)
{
   for(unsigned int i = 0; i != str.size() / 2; ++i)
   {
      if(str[i] != str[str.size() - 1 - i]) return false;
   }

   return true;
}

int main()
 {
    ofstream fout ("palsquare.out");
    ifstream fin ("palsquare.in");

    int n;
    fin>>n;

    for(int i = 1; i <= 300; ++i)
    {
       string sSquare;
       GetBaseValue(i * i, n, sSquare);
       if(IsPalindromic(sSquare))
       {
          string str;
          GetBaseValue(i, n, str);
          fout<<str<<" "<<sSquare<<endl;
       }
    }

    fin.close();
    fout.close();
    return 0;
}