USACO Palindromic Squares

简单的1-300递归搜索

主要的2处代码

/* 函数circle用于判断正整数n的d进制数表示形式是否是回文数 */
int circle(int n, int d)
{
int s=0,m=n;
while(m)
{
s=s*d+m%d;
m/=d;
}
return s==n;
}
//函数printd将正整数n由10进制转换为B进制
void printd( int n , int m )
{
char x;int i;
if( n < 0 )
{
fout<<'-';
n = -n;
}
if( n / m )
printd( n / m ,m);
if(n%m<10)
fout<<(n%m);
if(n%m>=10)
{
//fputc('A'+n%m-10,fout);
x='A'+n%m-10;
fout<<x;
// i=n%m-10;
// cout<<i;
// putchar('A');
}
}

Palindromic Squares
Rob Kolstad

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. 
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

===========================

AC代码:

/*
ID: jun41821
PROG: palsquare
LANG: C++
*/
#include <iostream>
#include <fstream>
#include <algorithm>
using namespace std;
ofstream fout ("palsquare.out");
ifstream fin ("palsquare.in");
/* 函数circle用于判断正整数n的d进制数表示形式是否是回文数 */
int circle(int n, int d)
{
int s=0,m=n;
while(m)
{
s=s*d+m%d;
m/=d;
}
return s==n;
}
//函数printd将正整数n由10进制转换为B进制
void printd( int n , int m )
{
char x;int i;
if( n < 0 )
{
fout<<'-';
n = -n;
}
if( n / m )
printd( n / m ,m);
if(n%m<10)
fout<<(n%m);
if(n%m>=10)
{
//fputc('A'+n%m-10,fout);
x='A'+n%m-10;
fout<<x;
// i=n%m-10;
// cout<<i;
// putchar('A');
}
}

int main()
{
int b,i,j;
fin>>b; //输入进制数

//在1 到 300 递归
//将i的平方换成B进制 判断是否为回文数 是就输出
for(i=1;i<=300;i++)
{
j=i*i;
if(circle(j,b))
{
printd(i,b);
fout<<' ';
printd(j,b);
fout<<endl;
}
}

}