USACO Dual Palindromes

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)

262728

这是一道挺有意思的题目，上次做了，且花了较长的时间做出来，今天又做了一遍，却又花了近两个小时，觉得自己比较笨，所以决定写下自己的思路。

题目要求我们判断一个数在2~10进制下，将一个数任意转换为其中的两种进制，看看在相应的进制下，是否为回文，若是，则输出该数。反之，继续判断，直到找到为止。

这里给出了进制转换的算法。在is_towpal函数里面。

下面是我的代码：

/*ID:lclqcsj1PROG:dualpalLANG:C++*/#include<iostream>#include<cstring>#include<fstream>using namespace std;
const int MAXN=100+5;
int is_towpal(int a){    int counter=0;    int base;    for(base=2; base<=10; base++)    {        char str[MAXN];        memset(str, 0, sizeof(str));        int t=a;        int i=0,k=0,r=0,j=0;        while((k=(t/base))!=t)        {            r=t%base;            str[i++]='0'+r;            t=k;        }        str[i]='\0';        char *pstart=str;        char *pend=str+i-1;        while(pstart<=pend)        {            if(*pstart==*pend)            {                pstart++;                pend--;            }            else                break;        }        if(pstart>pend)        {            counter++;        }        if(counter==2)            break;    }    if(counter==2)        return 1;    else        return 0;
}
int main(void){    ofstream fout ("dualpal.out");    ifstream fin ("dualpal.in");    int N,S;    fin>>N>>S;    S=S+1;    int T=1;    while(T<=N)    {        int flag=0;        flag=is_towpal(S);        if(flag)        {            T++;            fout<<S<<endl;            S++;        }        else            S++;    }    return 0;}