USACO section 1.2 Palindromic Squares

最新推荐文章于 2019-03-05 21:43:50 发布

原创最新推荐文章于 2019-03-05 21:43:50 发布 · 584 阅读

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本文介绍了一种算法，用于找出特定进制下平方后形成回文数的所有整数，并提供了C++实现代码及评测结果。

原文

Palindromic Squares
Rob Kolstad

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.

SAMPLE OUTPUT (file palsquare.out)

分析

本题内容非常简单，给定一个进制（2进制到20进制）之间，要求算出300（10进制）以内所有求平方结果在给定进制下是回文的数字，以给定的进制输出。例如给定10进制的话，得到的结果是

这道题的核心是不同进制之间数字的转换以及回文的判断，回文的判断非常简单，不同进制之间转换需要依据进制的定义来实现，例如，10进制下121表示 1*10^2+2*10^1+1*10^0，其他进制同理。

提交代码

/*
ID: Aaron Cai
PROG: palsquare
LANG: C++
*/
#include <fstream>
#include <algorithm>
#include <vector>
#include <string>
#include <math.h>
#include <iostream>
using namespace std;


string toBase(int a,int base)
{
	int length = log(double(a))/log(double(base))+1;
	string out="";
	int b = a;
	for (int i=0;i!=length;i++)
	{
		int w = b / pow(base*1.0,double(length-i-1));
		b = b - w * pow(base*1.0,double(length-i-1));

		if(w<10)
			out += char('0'+w);
		else
		{
			out += char('A'+w-10);
		}
	}

	return out;
}

bool isPalin(string str)
{
	int half = str.length()/2;
	bool same = true;
	for (int i=0;i!=half;i++)
	{
		same = same && str[i] == str[str.length()-i-1];
		if(!same)
			return false;
	}
	if(same)
		return true;
}

int main()
{
	ifstream fin("palsquare.in");
	int base; 
	fin >> base;


	ofstream fout("palsquare.out");
	
	//cout << toBase(2,2)<<endl;
	
	for (int i=1;i<=300;i++)
	{
		if (isPalin(toBase(i*i,base)))
		{
			fout << toBase(i,base) << " " << toBase(i*i,base) << endl;
		}
	}
	return 0;
}

提交结果

TASK: palsquare
LANG: C++

Compiling...
Compile: OK

Executing...
   Test 1: TEST OK [0.011 secs, 3496 KB]
   Test 2: TEST OK [0.003 secs, 3496 KB]
   Test 3: TEST OK [0.005 secs, 3496 KB]
   Test 4: TEST OK [0.005 secs, 3496 KB]
   Test 5: TEST OK [0.005 secs, 3496 KB]
   Test 6: TEST OK [0.005 secs, 3496 KB]
   Test 7: TEST OK [0.011 secs, 3496 KB]
   Test 8: TEST OK [0.008 secs, 3496 KB]

All tests OK.

官方参考答案

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#include <ctype.h>
#include <math.h>

/* is string s a palindrome? */
int
ispal(char *s)
{
    char *t;

    t = s+strlen(s)-1;
    for(t=s+strlen(s)-1; s<t; s++, t--)
    	if(*s != *t)
	    return 0;

    return 1;
}

/* put the base b representation of n into s: 0 is represented by "" */

void
numbconv(char *s, int n, int b)
{
    int len;

    if(n == 0) {
	strcpy(s, "");
	return;
    }

    /* figure out first n-1 digits */
    numbconv(s, n/b, b);

    /* add last digit */
    len = strlen(s);
    s[len] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"[n%b];
    s[len+1] = '\0';
}

void
main(void)
{
    char s[20];
    char t[20];
    int i, base;
    FILE *fin, *fout;

    fin = fopen("palsquare.in", "r");
    fout = fopen("palsquare.out", "w");
    assert(fin != NULL && fout != NULL);

    fscanf(fin, "%d", &base);
    for(i=1; i <= 300; i++) {
	numbconv(s, i*i, base);
	if(ispal(s)) {
	    numbconv(t, i, base);
	    fprintf(fout, "%s %s\n", t, s);
	}
    }
    exit(0);
}