USACA section 1.2.3 Palindromic Squares

1. 进制转换+回文判断+int与char之间相互转换；

以下是代码：

 

/*
ID: dollar4
PROG: palsquare
LANG: C++
*/
#include <iostream>
#include <fstream>
#include <string>
#include <algorithm>
#include <cstring>

using namespace std;
int base;
bool checkp(string str)
{
    int len = str.size();
    for (int i = 0; i < len; i++)
    {
        if (str[i] != str[len-1-i])
            return false;
    }
    return true;
}
string trans(int a)
{
    string str1 = "";
    int tmp;
    while (a)
    {
        tmp = a % base;
        if (tmp < 10)
            str1 += tmp + '0';
        else
            str1 += tmp - 10 + 'A';
        a /= base;
    }
    return str1;
}
string change(string str)
{
    string rlt = str;
    int len = str.size();
    for (int i = 0; i < len; i++)
        rlt[i] = str[len-1-i];
    return rlt;
}
int main()
{
    ofstream fout ("palsquare.out");
    ifstream fin ("palsquare.in");
    int i;
    string str1, str2;
    fin >> base;
    for (i = 1; i <= 300; i++)
    {
        str1 = trans(i * i);
        if (checkp(str1))
        {
            str2 = change(trans(i));
            fout << str2 << " " << str1 << endl;
        }
    }
    return 0;
}

以下是官方参考代码

#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);
}