1.2.4 & 1.2.5 Palindromic Squares & Dual Palindromes【水】

水题就不解释了，若干年前的代码，现在发现以前做过

/*
	ID: email_f1
	LANG: C++
	TASK: palsquare
*/
#include<string.h>
#include<stdio.h>
void tran(int m,int base,char *p)
{
	int r;
	while(m>0)
	{
		r=m%base;
		if(r<10)	*p=r+'0';
		else		*p=r+'A'-10;
		m=m/base;
		p++;
	}
	*p='\0';
}

int main()
{
    freopen("palsquare.in","r",stdin);
    freopen("palsquare.out","w",stdout);

	int i,b,m,j,k,q;
	char c[50],d[50];
	while ( scanf("%d",&b) !=EOF )
	for(i=1;i<=300;i++)
	{
		tran(i*i,b,c);
		for(k=strlen(c)-1,j=0;j<k;j++,k--)
			if(c[j]!=c[k])
			{
				break;
			}
		if(j>=k)
		{
			tran(i,b,d);
			for(q=strlen(d)-1;q>=0;q--)
				printf("%c",d[q]);
			printf(" ");

			for(q=0;q<strlen(c);q++)
				printf("%c",c[q]);
			putchar('\n');
		}
	}
	return 0;
}

/*
	ID: email_f1
	LANG: C++
	TASK: dualpal
*/
#include <cstdio>
#include <cmath>
#include <cstring>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
#define ll unsigned long long
#define INF 0x7FFFFFFF
#define eps 10^(-6)
#define Q_CIN ios::sync_with_stdio(false)
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define FOR( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define CLR( a , x ) memset ( a , x , sizeof (a) )
#define RE freopen("1.txt","r",stdin);
#define WE freopen("1.txt","w",stdout);
#define MOD 10009
#define NMAX 10002
#define min(a,b) ((a)>(b)?(b):(a))
#define max(a,b) ((a)<(b)?(b):(a))
string v="0123456789ABCDEF";

string change(int n,int base)
{
    string s1("");
    do
    {
        s1+=v[n%base];
        n/=base;
    }
    while(n>0);
    return s1;
}
bool ok(int n,int base)
{
    string s1=change(n,base);
    string s2=s1;
    reverse(s1.begin(),s1.end());
    if(s1==s2)
        return true;
    return false;
}
int main()
{
    freopen("dualpal.in","r",stdin);
    freopen("dualpal.out","w",stdout);

    int n,t;
    string str;
    while(cin>>t>>n)
    {
        int ans=0;
        FOR(j,n+1,2147483647)
        {
            int cnt=0;
            FOR(i,2,10)
            {
                if(ok(j,i))
                {
                    cnt++;
                }
                if(cnt>=2)
                {
                    cout<<j<<endl;
                    ans++;
                    break;
                }
            }
            if(ans==t)
                break;
        }
    }
    return 0;
}