USACO Training 1.3 Palindromic Squares

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本文介绍了一个程序，该程序能够找出在特定进制下其平方为回文数的所有整数（1≤N≤300）。回文数是指无论从前往后还是从后往前读都相同的数。文中提供了一个具体的实现方案，包括如何判断一个数是否为回文数以及如何将十进制数转换为目标进制表示。

Palindromes are numbers that read the same forwards as backwards. The number 12321 is a typical palindrome.

回文数是反向和正向读都是一样的数字，例如12321就是一个典型的回文数。

Given a number base B, 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)

#include <bits/stdc++.h> 
using namespace std; 
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;
}
void numbconv(char *s, int n, int b)
{
    int len;

    if(n == 0) {
	strcpy(s, "");
	return;
    } 
    numbconv(s, n/b, b); 
    len = strlen(s);
    s[len] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"[n%b];
    s[len+1] = '\0';
}

int main(){
    char s[20];
    char t[20];
    int i, base; 

    freopen("palsquare.in", "r", stdin);
    freopen("palsquare.out", "w", stdout);  
    cin>>base; 
    for(i=1; i <= 300; i++) {
    	numbconv(s, i*i, base);
    	if(ispal(s)) {
    	    numbconv(t, i, base); 
    	    cout<<t<<" "<<s<<endl;
    	}
    } 
}