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This is a explanation of this problem from USACO's training website. I have converted it to markdown. Please do not just copy code; you will not learn anything; at least type it out and understand so you can do it yourself in the future!

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)

10

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 1
2 4
3 9
11 121
22 484
26 676
101 10201
111 12321
121 14641
202 40804
212 44944
264 69696

CODE

Java

C++

Pascal

ANALYSIS

Russ Cox

We generate all the squares from 1 to 300 and check to see which are palindromes.

#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);
}

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