USACO 1.2 Palindromic Squares

這道題目就是個進制的轉換，簡單回文的判斷。

/*
	ID: acmerfi1
	PROG: palsquare
	LANG: C++
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX 100
#define maxNum 300

int base, num1[MAX], num2[MAX], squareI = 0, sumDigit1 = 0, sumDigit2 = 0;

int changeBase(int num[], int base, int t) //進制轉換 
{
	int i = 0;
	while(t)
	{
		num[i++] = t % base;
		t /= base;
	}
	return i; 		
}

int judge(int num[], int sumDigit) // 回文判斷 
{
	int flag = 0;
	for(int i = 0; i <= sumDigit/2; i++)
	{
		if(num[i] == num[sumDigit - i - 1]) flag = 1;
		else
		{
			flag = 0;
			break;	
		}
	}
	return flag;
}

int main()
{
	freopen("palsquare.in", "r", stdin);
	freopen("palsquare.out", "w", stdout);
	scanf("%d", &base);
	for(int i = 1; i <= maxNum; i++)
	{
	     memset(num1, 0, sizeof(num1));
	     memset(num2, 0, sizeof(num2));
		 squareI = i * i;
		 sumDigit1 = changeBase(num1, base, squareI);
		 sumDigit2 = changeBase(num2, base, i);
		 if(judge(num1, sumDigit1))
		 {
 			for(int j = sumDigit2 - 1; j >= 0; j--)
 			{
			 	if(num2[j] >= 0 && num2[j] <= 9) printf("%d", num2[j]);
			 	else printf("%c", num2[j] - 10 + 'A');
		    }
		    printf(" ");
		    for(int j = 0; j < sumDigit1; j++)
			{
				if(num1[j] >= 0 && num1[j] <= 9)
				{
					printf("%d", num1[j]);
				}
				else printf("%c", num1[j] - 10 + 'A');
			}
			printf("\n");
		 }
	}	
	return 0;
}