Dual Palindromes

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本文介绍了一种算法，用于找出大于指定数值的所有双基回文数，即在两个或以上不同进制下呈现回文特性的整数。通过逐个检查每个数在2到10进制下的表示是否为回文数来实现。

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Dual Palindromes
Mario Cruz (Colombia) & Hugo Rickeboer (Argentina)

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.

PROGRAM NAME: dualpal

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

代码：

/* 
 ID: jszhais1 
 PROG: dualpal 
 LANG: C++ 
 */  

//#include "stdafx.h"
//#include <iostream>
#include <fstream> 
using namespace std;

int main()
{
	ofstream fout("dualpal.out");  
    ifstream fin("dualpal.in"); 
	int zs_hexConversion(int num , const int &hex,char* zs_Num);
	bool isPalindromes(int length,const char* zs_Num);
	char zs_Num[2048];
	int N,S,count_s=0,length;
	fin>>N>>S;
	while(count_s<N)
	{
		S++;
		int count_n=0;
		for(int hex=2;count_n<2&&hex<=10;hex++)
		{
			length = zs_hexConversion(S,hex,zs_Num);
			if(isPalindromes(length,zs_Num))
			{count_n++;}

		}
		if(count_n>=2)
		{
			count_s++;
			fout<<S<<endl;
		}
	}

	return 0;
}

int zs_hexConversion(int num , const int &hex,char* zs_Num)
{
	int i=0,tempN=0;
	for(;num!=0;i++)
	{
		tempN = num%hex;
		num = num/hex;
		zs_Num[i]='0'+tempN;
	}
	zs_Num[i]='\0';
	return i;
}

bool isPalindromes(int length,const char* zs_Num)
{
	length--;
	for(int i=0;i<length;i++,length--)
	{
		if(zs_Num[i]!=zs_Num[length])
			return false;
	}
	return true;
}

思路：

考察进制转化，数据量不大故采用暴力循环法即可。

结果：

USER: jim zhai [jszhais1]
TASK: dualpal
LANG: C++

Compiling...
Compile: OK

Executing...
Test 1: TEST OK [0.000 secs, 3356 KB]
Test 2: TEST OK [0.000 secs, 3356 KB]
Test 3: TEST OK [0.000 secs, 3356 KB]
Test 4: TEST OK [0.000 secs, 3356 KB]
Test 5: TEST OK [0.000 secs, 3356 KB]
Test 6: TEST OK [0.000 secs, 3356 KB]
Test 7: TEST OK [0.000 secs, 3356 KB]

All tests OK.
Your program ('dualpal') produced all correct answers! This is your
submission #2 for this problem. Congratulations!

Here are the test data inputs:

------- test 1 ----
5 1
------- test 2 ----
9 10
------- test 3 ----
15 9900
------- test 4 ----
10 90
------- test 5 ----
12 125
------- test 6 ----
12 1900
------- test 7 ----
8 500
Keep up the good work!
Thanks for your submission!

答案:

Dual Palindromes
Russ Cox

Dual palindromes are actually very common, a fact we can test by writing a program such as this one.

Since they are very common, we can just use a brute force search to test all numbers bigger than s until we find enough dual palindromes.

How do we know they are common enough? Write the brute force program (which is very simple and thus not much effort) and check.

This reasoning is a little circular, but if we had been wrong and ended up needing a more clever and more efficient algorithm, we would have this brute force version to test against.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.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 "" */
char*
numbconv(char *s, int n, int b)
{
    int len;

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

    /* 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';
    return s;
}

/* is number n a dual palindrome? */
int
isdualpal(int n)
{
    int i, j, npal;
    char s[40];

    npal = 0;
    for(i=2; i<=10; i++)
	if(ispal(numbconv(s, n, i)))
	    npal++;

    return npal >= 2;
}

void
main(void)
{
    FILE *fin, *fout;
    int n, s;

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

    fscanf(fin, "%d %d", &n, &s);

    for(s++; n>0; s++) {
	if(isdualpal(s)) {
	    fprintf(fout, "%d\n", s);
	    n--;
	}
    }

    exit(0);
}

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