USACO--Palindromic Squares

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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. NOTE WELL THAT BOTH INTEGERS ARE IN BASE B!

SAMPLE OUTPUT (file palsquare.out)

1 12 43 911 12122 48426 676101 10201111 12321121 14641202 40804212 44944264 69696

这道题的题意是让你输入进制，然后输出1-300中在某个数的平方这个进制下是回文数的话，就在这个进制下输出这个数本身和这个数的平方

思路：从1-300 每个数转换到对应的进制，如果是回文，就把数本身也转换一下，然后输出

代码：

#include<cstdio>#include<cstring>#include<iostream>#define maxn 1500using namespace std;char hash_arr[20]= {'0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F','G','H','I'};  //对应的进制转换int base;int is_dual(char* s,int len)     //判断是否为回文{    bool ok=true;    for(int i=0; i<=len; i++)    {        if(s[i]!=s[len])        {            ok=false;break;        }    }    return ok;}int base_trans(int n,char* tr)  //将数的平方转换成 对应的进制，然后去判断是否回文{    int len=0;    while(n)    {        tr[len++]=hash_arr[n%base];        n/=base;    }    return len;    //返回转换后的字符串的长度}int main(){    scanf("%d",&base);    char tr[maxn],num[maxn];    for(int i=0; i<=300; i++)    {        memset(tr,0,sizeof(tr));        memset(num,0,sizeof(num));        int lensqu=base_trans(i*i,tr);  //lensqu是平方数转换后的长度        if(is_dual(tr,len-1))        {            int lennum=base_trans(i,num);  //lennum是数本身转换后的长度            for(int j=lennum-1; j>=0; j--)                   printf("%c",num[j]);            printf(" ");            for(int j=lensqu-1; j>=0; j--)                printf("%c",tr[j]);            printf("\n");        }    }    return 0;}

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